Rest and Motion Kinematics

Concept Of Physics

H C Verma

1   A man has to go $50$ m due north, $40$ m due east and $20$ m due south to reach a field. (a) What distance he has to walk to reach the field ? (b) What is his displacement from his house to the field ?

Solution :

a) Distance travelled = $50$ + $40$ + $20$ = $110$ m

b) $AF$ = $AB$ - $BF$ = $AB$ - $DC$ = $50$ - $20$ = $30$ M $\\$ His displacement is $AD$

$AD$ = $\sqrt{AF^2-DF^2}$ = $\sqrt{30^2+40^2}$ = $50$ M

In $\Delta{AED}$ tan$\theta$ = $\frac{DE}{AE}$ = $\frac{30}{40}$ = $\frac{3}{4}$

$\Rightarrow$ $\theta$ = $tan^{-1}$ ($\frac{3}{4}$)

His displacement from his house to the field is $50$ m , $\\$ $tan^{-1}$($\frac{3}{4}$) north to east

2   A particle starts from the origin, goes along the X-axis to the point $(20 m, 0)$ and then returns along the same line to the point $(-20 m, 0)$. Find the distance and displacement of the particle during the trip.

Solution :

O $\rightarrow$ Starting point origin. $\\$ i ) Distance travelled = $20$ + $20$ + $20$ = $60$ m $\\$ ii ) Displacement is only OB = $20$ m in the negative direction. $\\$ Displacement $\rightarrow$ Distance between final and initial position.

3   It is $260$ km from Patna to Ranchi by air and $320$ km by road. An aeroplane takes $30$ minutes to go from Patna to Ranchi whereas a delux bus takes $8$ hours. $\\$(a) Find the average speed of the plane. $\\$(b) Find the average speed of the bus. $\\$ (c) Find the average velocity of the plane. $\\$(d) Find the average velocity of the bus.

Solution :

d) Straight path distance between plane to ranchi is equal to the displacement of bus.$\\$ $\therefore$ velocity = $\vec{V}_{avg}$ = $\frac{260}{8}$ = 32.5$\frac{km}{hr}$

a) $\ V_{avg}$ of plane ($\frac{Distance}{Time}$) = $\frac{260}{0.5}$ = 520$\frac{km}{hr}$

c) Plane goes in straight path $\\$ velocity = $\vec{V}_{avg}$ = $\frac{260}{0.5}$ = 520$\frac{km}{hr}$

b) $\ V_{avg}$ of bus = $\frac{320}{8}$ = 40$\frac{km}{hr}$

4   When a person leaves his home for sightseeing by his car, the meter reads $12352$ km. When he returns home after two hours the reading is $12416$ km.$\\$ (a) What is the average speed of the car during this period ? $\\$(b) What is the average velocity ?

Solution :

a ) Total distance covered $12416$ - $12352$ = $64$ km in $2$ hours. $\\$ speed = $\frac{64}{2}$ = 32$\frac{km}{hr}$

b) As he returns to his house , the displacement is zero. $\\$ Velocity = $\frac{displacement}{time}$ = 0(zero)

5   An athelete takes $2.0$ s to reach his maximum speed of $18.0$ km/h. What is the magnitude of his average acceleration ?

Solution :

Initial velocity u = 0 ( $\therefore$ starts from rest) $\\$ final velocity v = 18 $\frac{km}{hr}$ = 5 sec.

(i.e max velocity) $\\$ The interval t = 2 sec $\\$ $\therefore$ acceleration = $a_{avg}$ = $\frac{v-u}{t}$ = $\frac{5}{2}$ = 2.5$\frac{m}{s^2}$

6   The speed of a car as a function of time is shown in figure (3-E1). Find the distance travelled by the car in $8$ seconds and its acceleration.

Solution :

In the interval 8 sec the velocity changes from 0 to 20 $\frac{m}{s}$ $\\$ average acceleration = $\frac{20}{8}$ = 2.5 $\frac{m}{s^2}$ ($\frac{change in velocity}{time}$)

Distance Travelled $s=ut+\frac{1}{2}at^2$ $\\$ $\Rightarrow 0+\frac{1}{2}(2.5)8^2=80m$.

Distance Travelled $s=ut+\dfrac{1}{2}at^2$

$\implies 0+\dfrac{1}{2}(2.5)8^2=80m$

7   The acceleration of a cart started at $t = 0$, varies with time as shown in figure (3-E2). Find the distance travelled in $30$ seconds and draw the position-time graph.

Solution :

In $1^{st}$ 10 sec $S_{1}$ = ut + $\frac{1}{2}$a$t^{2}$ $\Rightarrow$ 0 + ( $\frac{1}{2} \times 5 \times 10^2$) = 250 ft.

At 10 sec v = u + at = 0 + 5 $\times$ 10 = 50 $\frac{ft}{sec}$ .

$\therefore$ From 10 to 20 sec ($\Delta$t = 20 - 10 = 10 sec) it moves with uniform velocity 50 $\frac{ft}{sec}$ .

Distance $S_2 = 50 \times 10 = 500 ft.$

Between 20 sec to 30 sec Acceleration is constant i.e -5$\frac{ft}{s^2}$ . At 20 sec velocity is 50$\frac{ft}{sec}$.

t = 30 - 20 = 10 sec.

$S_3 = ut + \frac{1}{2}at^2 = 50 \times 10 + (\frac{1}{2})(-5)(10)^2$ = 250 m

Total distance travelled in 30 sec = $S_1$ + $S_2$ + $S_3$ = 250 + 500 + 250 = 1000 ft.

8   Figure $(3-E3)$ shows the graph of velocity versus time for a particle going along the X-axis. Find $\\$ (a) the acceleration, $\\$ (b) the distance travelled in 0 to 10 s and $\\$ (c) the displacement in 0 to 10 s.

Solution :

a) Initial velocity u = 2 $\frac{m}{s}$. $\\$ Final velocity v = 8 $\frac{m}{s}$. $\\$ time = 10 sec, $\\$ acceleration = $\frac{v-u}{at}$ = $\frac{8-2}{10}$ = 0.6 $\frac{m}{s^2}$

b) $v^2$ - $u^2$ = 2aS $\\$ $\Rightarrow$ Distance S = $\frac{v^2-u^2}{2a}$ = $\frac{8^2-2^2}{2 \times 0.6}$ = 50 m.

c) Displacement is same as distance travelled .

Displacement = 50 m.

9   Figure (3-E4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time. Find $\\$ (a) the average velocity during $0$ to $10$ s, $\\$ (b) instantaneous velocity at $2$, $5$, $8$ and $12s$.

Solution :

b) At 2 sec it is moving with uniform velocity $\frac{50}{2.5}$ = 20 $\frac{m}{s}$.

a) Displacement in 0 to 10 sec is 1000 m.

time = 10 sec.

$v_{avg}$ = $\frac{s}{t}$ = $\frac{100}{10}$ = 10 $\frac{m}{s}$

b) At 2 sec it is moving with uniform velocity $\frac{50}{2.5}$ = 20 $\frac{m}{s}$

At 2 sec . $V_{inst}$ = 20 $\frac{m}{s}.$

At 5 sec it is at rest.

$V_{inst}$ = $zero$

At 8 sec it is moving with uniform velocity 20 $\frac{m}{s}$

$V_{inst}$ = 20 $\frac{m}{s}$

At 12 sec velocity is negative as it move toward initial position . $V_{inst}$ = -20 $\frac{m}{s}$

10   From the velocity—time plot shown in figure (3-E5), find the distance travelled by the particle during the first $40$ seconds. Also find the average velocity during this period.

Solution :

Distance in first 40 sec is , $\Delta{OAB}$ + $\Delta{BCD}$

= $\frac{1}{2} \times 5 \times 20$ + $\frac{1}{2} \times 5 \times 20$ = 100 m.

Average velocity is 0 as the displacement is zero.

11   Figure (3-E6) shows $x-t$ graph of a particle. Find the time $t$ such that the average velocity of the particle during the period $0$ to $t$ is zero.

Solution :

Consider the point B , at t = 12 sec $\\$ At t = 0 ; s = 20 m $\\$ and t = 12 sec ; s = 20m $\\$ So for time interval 0 to 12 sec. $\\$ Change in displacement is zero. $\\$ So average velocity = $\frac{displacement}{time}$ = 0 $\\$ $\therefore$ The time is 12 sec.

12   A particle starts from a point A and travels along the solid curve shown in figure $(3-E7)$. Find approximately the position B of the particle such that the average velocity between the positions A and B has the same direction as the instantaneous velocity at B.

Solution :

We can see that $\vec{AB}$ is along $\vec{BC}$ i.e they are in same direction.

$V_{avg}$ = $\frac{displacement}{time}$ = $\frac{(\vec{AB})}{t}$ $\\$ t= time

The point is B (5m , 3m).

At position B instantaneous velocity has direction along $\vec{BC}$. For average velocity between A and B.

13   An object having a velocity $4.0$ m/s is accelerated at the rate of $1.2$ m/s2 for $5.0$ s. Find the distance travelled during the period of acceleration.

Solution :

u = 4 $\frac{m}{s}$ , a = 1.2 $\frac{m}{s^2}$ , t = 5 sec $\\$ Distance = s = $ut$ + $\frac{1}{2}at^2$

= $4(5) + \frac{1}{2}(1.2)5^2$ = $35$ m.

14   A person travelling at $43.2$ km/h applies the brake giving a deceleration of $6.0$ $m/s2$ to his scooter. How far will it travel before stopping ?

Solution :

Initial velocity u = $43.2$ $\frac{km}{hr}$ = $12$ $\frac{m}{s}$ $\\$ u = $12$ $\frac{m}{s}$ , v = 0 $\\$ a = $-6$ $\frac{m}{s^2}$ (deceleration)

Distance S = $\frac{v^2-u^2}{2(-6)}$ = $12$ m

15   A train starts from rest and moves with a constant acceleration of $2.0$ $m/s2$ for half a minute. The brakes are then applied and the train comes to rest in one minute. Find $\\$(a) the total distance moved by the train,$\\$ (b) the maximum speed attained by the train and $\\$ (c) the position(s) of the train at half the maximum speed

Solution :

$v' = 0 , t = 60$ sec ( 1 min)

$Initial$ $velocity$ $u = 0$ $\\$ $Acceleration$ $a = 2$ $\frac{m}{s^2}$. $Let$ $final$ $velocity$ $be$ v ( before applying breaks ) $\\$ t = $30$ $sec$ $\\$ $v = u + at \Rightarrow 0 + 2 \times 30 = 60$ $\frac{m}{s}$

b) The maximum speed attained by train v = $60$ $\frac{m}{s}$

When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 $\frac{m}{s}.$

when breaks are applied u' = $60 \frac{m}{s}$

Distance = $\frac{v^2-u^2}{2a}$ = $\frac{30^2-60^2}{2(-1)}$ = $1350$ m $\\$ Position is $900$ + $1350$ = $2250$ = $2.25$ $km$ from starting point.

Declaration $a'$ = $\frac{(v-u)}{t}$ == $\frac{(0-60)}{60}$ = -1 $\frac{m}{s^2}$. $\\$ $S_2$ = $\frac{(v')^2-(u')^2}{2a'}$ = $1800$ m $\\$ Total S = $S_1$ + $S_2$ = $1800$ + $900$ = $2700$ m = $2.7$ km.

Distance S = $\frac{v^2-u^2}{2a}$ = $\frac{30^2-0^2}{2\times2}$ = $225$ m from starting point

a) $S_1 = ut + \frac{1}{2}at^2 = 900 m$

c) Half the maximum speed = $\frac{60}{2}$ = $30$ $\frac{m}{s}$

$\therefore$ u = $60$ $\frac{m}{s}$ , v = $30$ $\frac{m}{s}$ , a = -1 $\frac{m}{s^2}$

16   A bullet travelling with a velocity of $16$ $m/s$ penetrates a tree trunk and comes to rest in $0.4$ $m$. Find the time taken during the retardation.

Solution :

$u = 16 \frac{m}{s}$ (initial) , $v =0,$ $s =0.4m$

Time = $t$ = $\frac{v-u}{a}$ = $\frac{0-16}{-320}$ = $0.05$ $sec$.

Deceleration $a$ = $\frac{v^2-u^2}{2s}$ = $-320$ $\frac{m}{s^2}$.

17   A bullet going with speed $350$ $m/s$ enters a concrete wall and penetrates a distance of $5.0$ $cm$ before coming to rest. Find the deceleration.

Solution :

$u$ = $350$ $\frac{m}{s}$ , $s$ = $5$ $cm$ = $0.05m$ , $v=0$

Deceleration is $12.2 \times 10^5$ $\frac{m}{s^2}.$

Deceleration = $a$ = $\frac{v^2-u^2}{2a}$ = $\frac{0-(350)^2}{2\times0.05}$ = $-12.2 \times 10^5$ $\frac{m}{s^2}$.

18   A particle starting from rest moves with constant acceleration. If it takes $5.0$ $s$ to reach the speed $18.0$ $km/h$ find $\\$(a) the average velocity during this period, and $\\$(b) the distance travelled by the particle during this period.

Solution :

$u = 0$ , $v = 18\frac{km}{hr}$ = $5$ $\frac{m}{s}$ , $t$ = $5$ $sec$

a) Average velocity $V_{avg} = \frac{(12.5)}{5} = 2.5\frac{m}{s}$. $\\$ b) distance travelled is $12.5 m$

$a$ = $\frac{v-u}{t}$ = $\frac{5-0}{5}$ = $1$ $\frac{m}{s^2}.$ $\\$ $s$ = $ut + \frac{1}{2}at^2 = 12.5m$

19   A driver takes $0.20$ $s$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of $54$ $km/h$ and the brakes cause a deceleration of $6.0$ $m/s^2$, find the distance travelled by the car after he sees the need to put the brakes on

Solution :

In reaction time the body moves with the speed $54$ $\frac{km}{hr}$ = $15$ $\frac{m}{sec}$ (constant speed) $\\$ Distance travelled in this time $S_1= 15\times0.2 = 3m$. $\\$ When brakes are applied , $\\$ $u$ = 15 $\frac{m}{s}$, $v$ = 0, $a = -6\frac{m}{s^2}$ (deceleration) $\\$ $S_2 = \frac{v^2-u^2}{2a} = \frac{0-15^2}{2(-6)} = 18.75m$

Total distance $s = s_1 +s_2 = 3+18.75 = 21.75 = 22m$.

20   A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of $72$ $km/h$. The jeep follows it at a speed of $90$ $km/h$, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike ?

Solution :

$V_P = 90\frac{km}{h} = 25\frac{m}{s}.$ $\\$ $V_C= 72\frac{km}{h} = 20\frac{m}{s}.$ $\\$ In 10 sec culprit reaches at point B from A. $\\$ Distance converted by culprit $S = vt=20\times10= 200 m$ $\\$ At time $t= 10 sec$ the police jeep is $200m$ behind the culprit. $\\$ Time = $\frac{s}{v} = \frac{200}{5} =40s.$ (Relative velocity is considered).

In $40s$ the police jeep will move from $A$ to a distance $S$, where $S =vt=25\times40=1000m=1.0km $ away. $\\$ $\therefore$ The jeep will catch up the bike , $1km$ far from the turning.

21   A car travelling at $60$ $km/h$ overtakes another car travelling at $42$ $km/h$. Assuming each car to be $5.0$ $m$ long, find the time taken during the overtake and the total road distance used for the overtake.

Solution :

In $2$ $sec$ the $1^{st}$ car moved = $16.6 \times2=33.2$ $\\$ H also covered its own length $5$ $m$ $\\$ $\therefore$ Total road distance used for the overtake = $33.2+5=38m$

$V_1 = 60\frac{km}{hr} = 16.6 \frac{m}{s}.$ $\\$ $V_2 = 42\frac{km}{hr} = 11.6 \frac{m}{s}.$ $\\$ Relative velocity between the cars = $(16.6-11.6)= 5\frac{m}{s}.$ $\\$ Distance to be travelled by first car is $5+t=10m.$ $\\$ $Time = t=\frac{s}{v} = \frac{0}{5}=2 sec$ to cross the $2^{nd}$ car.

22   A ball is projected vertically upward with a speed of $50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to reach the maximum height, $\\$(c) the speed at half the maximum height. Take g = $10$ $m/s^2$

Solution :

$\Rightarrow$v = $\sqrt{(u^2+2as)} = \sqrt{(50^2+2(-10)(62.5))} = 35\frac{m}{s}$.

$u$ = $50 \frac{m}{s} , g=-10\frac{m}{s^2}$ When moving upward , $v =0 $ (at highest point)

b) $ t=\frac{(v-u)}{a} = \frac{(0-50)}{-10} = 5 sec$ $\\$ c) $s' = \frac{125}{2} = 62.5m$, $u=50\frac{m}{s}, a = -10\frac{m}{s^2}, v^2-u^2=2as$ $\\$

a) $S= \frac{v^2-u^2}{2a}=\frac{0-50^2}{2(-10)} = 125m$ $\\$ maximum height reached = 125 m

23   A ball is projected vertically upward with a speed of $50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to reach the maximum height, $\\$(c) the speed at half the maximum height. Take g = $10$ $m/s^2$

Solution :

24   A ball is projected vertically upward with a speed of $50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to reach the maximum height, $\\$(c) the speed at half the maximum height. Take g = $10$ $m/s^2$

Solution :

25   A ball is projected vertically upward with a speed of $50$ $m/s$. Find $\\$(a) the maximum height, $\\$(b) the time to reach the maximum height, $\\$(c) the speed at half the maximum height. Take g = $10$ $m/s^2$

Solution :

26   A ball is dropped from a balloon going up at a speed of $7$ $m/s$. If the balloon was at a height $60$ $m$ at the time of dropping the ball, how long will the ball take in reaching the ground ?

Solution :

taking positive sign $t$ = $\frac{7+35.34}{10}$ = $4.2 sec$ ( $\therefore$ t $\neq$ -ve)

Initially the ball is going upward $\\$ $u =-7\frac{m}{s},$ $s=60m$ , $a =g=10\frac{m}{s^2}$ $\\$ $s=ut+\frac{1}{2}at^2$ $\Rightarrow$ $60=-7t+\frac{1}{2}10t^2$

$t$=$\frac{7\pm\sqrt{49-4.5(-60)}}{2\times5}$ = $\frac{7\pm35.34}{10}$

Therefore , the ball will take 4.2 sec to reach the ground.

$\Rightarrow$ $5t^2-7t-60=0$

27   A stone is thrown vertically upward with a speed of $28$ $m/s$. $\\$(a) Find the maximum height reached by the stone. $\\$(b) Find its velocity one second before it reaches the maximum height. $\\$(c) Does the answer of part (b) change if the initial speed is more than $28$ $m/s$ such as $40$ $m/s$ or $80$ $m/s$ ?

Solution :


$u =28\frac{m}{s},$ $v=0$ , $a=g=-9.8\frac{m}{s^2}$

$\therefore$ The velocity is 9.87 $\frac{m}{s}.$

b) time $t=\frac{v-u}{a} = \frac{0-28}{-9.8}=2.85$


a) $S=\frac{v^2-u^2}{2a} =\frac{0-28^2}{2((9.8)} =40m$

c) No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is $g=9.8\frac{m}{s^2}$ remains same. for initial velocity more than $28 \frac{m}{s}$ max height increases.

28   A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the $3rd$, $4th$ and $5th$ ball when the $6th$ ball is being dropped.

Solution :

$S_3 = ut+\frac{1}{2}at^2=0+\frac{1}{2}(9.8)3^2=4.9m$ below the top.

$S_2 = 0+\frac{1}{2}gt^2=\frac{1}{2}(9.8)2^2=19.6m$ below the top (u=0).

For $3^{rd}$ ball $t=3sec$ $\\$

$S_3 = ut+\frac{1}{2}at^2=0+\frac{1}{2}(9.8)1^2=4.9m$ below the top.

For $4^{th}$ ball $t=2sec$ $\\$

For every ball , $u=0 , a=g=9.8\frac{m}{s^2}$ $\\$ $\therefore$ $4^{th}$ ball move for $2$ $sec$, $5^{th}$ ball $1 sec$ and $3^{rd}$ ball $3sec$ when $6^{th}$ ball is being dropped.

For $5^{th}$ ball $t=1sec$ $\\$

29   A healthy youngman standing at a distance of $7$ $m$ from a $11.8$ $m$ high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height $(1.8 m)$ ?

Solution :

At point B( i.e. over $1.8m$ from ground) the kid should be catched. $\\$ For kid initial velocity u=0 $\\$ Acceleration = $9.8 \frac{m}{s^2}$ $\\$ Distance $S=11.8-1.8=10m$ $\\$ $S=ut+\frac{1}{2}at^2 \Rightarrow 10=0+\frac{1}{2}(9.8)t^2$ $\\$ $\Rightarrow t^2=2.04 \Rightarrow t=1.42.$

In this time the man has to reach at the bottom of the building. $\\$ Velocity $\frac{s}{t} = \frac{7}{1.42}=4.9\frac{m}{s}.$

30   An NCC parade is going at a uniform speed of $6$ $km/h$ through a place under a berry tree on which a bird is sitting at a height of $12.1$ $m$. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform ?

Solution :

$\frac{my \quad name}{name}$

Let the time of fall be 't' . initial velocity u=0 $\\$ Acceleration $a= 9.8\frac{m}{s^2}$ Distance $S= \frac{12}{1}m$ $\\$ $\therefore S=ut+\frac{1}{2}at^2$ $\\$ $\Rightarrow 12.1=0+\frac{1}{2}(9.8)\times$ $t^2$

For cadet velocity = 6 $\frac{km}{hr}=1.66\frac{m}{sec}$ $\\$ Distance = $vt=1.57\times1.66=2.6m$. $\\$ The cadet , $2.6$ m away from tree will receive the berry on his uniform.

$\Rightarrow$ $t^2=\frac{12.1}{4.9}=2.46 \Rightarrow t=1.57sec$

31   A ball is dropped from a height. If it takes $0.200$ $s$ to cross the last $6.00$ m before hitting the ground, find the height from which it was dropped. Take $g$ = $10$ $m/s^2$.

Solution :

For last $6\ m\ distance\ travelled\ s = 6m\ , u=?$ $\\$ $t=0.2\ sec,\ a=g=9.8\ \frac{m}{s^2}$

$S=\frac{v^2-u^2}{2a} = \frac{29^2-0^2}{2\times 9.8}=42.05\ m.$

For last $6\ m\ distance\ travelled\ s = 6m$

$S=ut+\frac{1}{2}at^2 \Rightarrow 6=u(0.2)+4.9\times0.04$ $\\$ $\Rightarrow u=\frac{5.8}{0.2} =29\ \frac{m}{s}.$ $\\$ $For\ distance\ x\ ,\ u=0,\ v=29\ \frac{m}{s},\ a =g=9.8\ \frac{m}{s^2}$ $\\$

$Total\ distance=42.05+6=48.05=48\ m.$

32   A ball is dropped from a height of $5$ $m$ onto a sandy floor and penetrates the sand up to $10$ $cm$ before coming to rest. Find the retardation of the ball in sand assuming it to be uniform

Solution :

$Consider\ the\ motion\ of\ ball\ from\ A\ to\ B. \\ B \rightarrow just\ above\ the\ sand\ (just\ to\ penetrate)\\ u=0,\ a=9.8\ \frac{m}{s^2},\ s=5\ m.\\ S=ut+\frac{1}{2}at^2\\ \Rightarrow 5=0+\frac{1}{2}(9.8)t^2\\ \Rightarrow t^2=\frac{5}{4.9}=1.02\Rightarrow t=1.01.\\ \therefore velocity\ at\ B,\ v=u+at=9.8\times 1.01\ (u=0)\ =9.89\ \frac{m}{s}.$

$From\ motion\ of\ ball\ in\ sand$ $\\$ $u_1=9.89\ \frac{m}{s},\ v_1=0,\ $ $\\$ $a=?,\ s=10\ cm=0.1m.$ $\\$ $a=\frac{v_{1}^2-u_{1}^2}{2s}=\frac{0-(9.89)^2}{2\times 0.1}= -490\ \frac{m}{s^2}\\ The\ retardation\ in\ sand\ is\ 490\ \frac{m}{s^2}.$

33   An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is $6$ $ft$ above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in $1$ $second$. Calculate from these data the acceleration of the elevator

Solution :

$For\ elevator\ and\ coin\ u=0$ $\\$ $As\ the\ elevator\ descends\ downward\ with\ acceleration\ a'\ (say)$ $\\$ $The\ coin\ has\ to\ move\ more\ distance\ than\ 1.8m\ to\ strike\ the\ floor.\ Time\ taken\ t=1\ sec.$ $\\$ $S_c=ut+\frac{1}{2}a't^2=0+\frac{1}{2}g(1)^2=\frac{1}{2}g$ $\\$ $S_e=ut+\frac{1}{2}at^2=u+\frac{1}{2}a(1)^2=\frac{1}{2}a$ $\\$ $Total\ distance\ covered\ by\ coin\ is\ given\ by\ =1.8+\frac{1}{2}a=\frac{1}{2}g$ $\\$ $\Rightarrow1.8+\frac{a}{2}=\frac{9.8}{2}=4.9\\ \Rightarrow=6.2\frac{m}{s^2}=6.2\times3.28=20.34\ \frac{ft}{s^2}.$

34   A ball is thrown horizontally from a point $100$ $m$ above the ground with a speed of $20$ $m/s$. Find $\\$(a) the time it takes to reach the ground, $\\$(b) the horizontal distance it travels before reaching the ground, $\\$(c) the velocity (direction and magnitude) with which it strikes the ground.

Solution :

$=\sqrt{\frac{2\times100}{9.8}}=4.51\ sec.$ $\\$ $b)\ Horizontal\ range\ x=ut = 20\times 4.5=90\ m.$ $\\$ $c)\ Horizontal\ velocity\ remains\ constant\ through\ out\ the\ motion$ $\\$ $At\ A,\ V=20\ \frac{m}{s}\\ AV_y=u+at=0+9.8\times4.5=44.1\ \frac{m}{s}.\\ Resultant\ velocity\ V_r=\sqrt{(44.1)^2+20^2}=48.42\ \frac{m}{s}.$ $\\$ $tan\beta=\frac{ V_y}{V_x}=\frac{44.1}{20}=2.205$

$The\ ball\ strikes\ the\ ground\ with\ a\ velocity\ 48.42\ \frac{m}{s}\ at\ an\ angle\ 66^{\circ}\ with\ horizontal.$

$It\ is\ a\ case\ of\ projectile\ fired\ horizontally\ from\ a\ height$ $\\$ $h=100\ m,\ g=9.8\ \frac{m}{s^2}$ $\\$ $a)\ Time\ taken\ to\ reach\ the\ ground\ t\ =\sqrt{(\frac{2h}{g})}$

$\Rightarrow \beta=tan^{-1}(2.205) = 60^{\circ}$

35   A ball is thrown at a speed of $40$ $m/s$ at an angle of $60°$ with the horizontal. Find $\\$(a) the maximum height reached and $\\$(b) the range of the ball. $\\$Take $g$ = $10$ $m/s^2$

Solution :

$u=40\ \frac{m}{s},\ a=g=9.8\ \frac{m}{s^2},\ \theta=60^{\circ}\ Angle\ of\ projection.\\ a)\ Maximum\ height\ h=\frac{u^{2}sin^{2}\theta}{2g}=\frac{40^{2}(sin60^\circ)^{2}}{2\times10}=60\ m$

$b)\ Horizontal\ range\ X=\frac{(u^2sin2\theta)}{g}=\frac{(40^2sin2(60^\circ))}{10}=80\sqrt{3}m$

36   In a soccer practice session the football is kept at the centre of the field $40$ $yards$ from the $10$ $ft$ high goalposts. A goal is attempted by kicking the football at a speed of $64$ $ft/s$ at an angle of $45°$ to the horizontal. Will the ball reach the goal post ?

Solution :


In time $2.65,$ the ball travels horizontal distance $120\ ft\ (40\ yd)$ and vertical height $7.08\ ft$ which is less 10 ft. The ball will reach the goal post.

$\Rightarrow=t=\frac{x}{ucos\theta}=\frac{120}{64cos45^\circ}=2.65\ sec$

$=7.08\ ft\ which\ is\ less\ than\ the\ height\ of\ goal\ post.$

$g=9.8\ \frac{m}{s^2},\ 32.2\ \frac{ft}{s^2}\ ;\ 40\ yd=120\ ft\\ horizontal\ range\ x=120\ ft,\ u=64\ \frac{ft}{s},\ \theta = 45^\circ\\ We\ know\ that\ horizontal\ range\ X = u\ cos{\theta}t$

37   A popular game in Indian villages is goli which is played with small glass balls called golis. The goli of one player is situated at a distance of $2.0$ $m$ from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is $19.6\ cm$ from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationary goli without falling on the ground earlier ?

Solution :

$The\ goli\ moves\ like\ a\ projectile\\ Here\ h=0.196\ m.\\ Horizontal\ distance\ X=2\ m.\\ Acceleration\ g=9.8\ \frac{m}{s^2}.\\ Time to reach the ground i.e. \\ t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times0.196}{9.8}}=0.2\ sec.\\ Horizontal\ velocity\ with\ which\ it\ is\ projected\ be\ u.\\ \therefore x=ut\\ \Rightarrow\ u=\frac{x}{t}=\frac{2}{0.2}=10\ \frac{m}{s}.$

38   Figure (3-E8) shows a $11.7$ $ft$ wide ditch with the approach roads at an angle of $15°$ with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch ? Assume that the length of the bike is $5$ $ft$, and it leaves the road when the front part runs out of the approach road.

Solution :

$Horizontal\ range\ X=11.7+5=16.7\ ft\ covered\ by\ te\ bike.\\ g=9.8\ \frac{m}{s^2}=32.2\ \frac{ft}{s^2}.$ $\\$ $y=xtan\theta-\frac{gx^2sec^2\theta}{2u^2} \Rightarrow\ u^2=\frac{gx^2sec^2\theta}{2xtan\theta}=\frac{gx}{2sin\theta (cos\theta)}=\frac{gx}{sin2\theta}$ $\\$ $\Rightarrow u=\sqrt{\frac{(32.2)(16.7)}{\frac{1}{2}}}\ (beacuse\ sin30^\circ=\frac{1}{2})$ $\\$ $\Rightarrow u =32.79\ \frac{ft}{s}= 32\ \frac{ft}{s}.$

39   A person standing on the top of a cliff $171$ $ft$ high has to throw a packet to his friend standing on the ground $228$ $ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15.0$ $ft/s$, how short will the packet fall ?

Solution :

$tan\theta=\frac{171}{228}\Rightarrow \theta=tan^{-1}(\frac{171}{228})\\ The\ motion\ of\ projectile\ (i.e.\ the\ packed)\ is\ from\ A.\ Taken\ references\ axis\ at\ A.\\ \therefore\theta=-37^\circ\ as\ u\ is\ below\ x-axis.\\ u=15\ \frac{ft}{s},\ g=32.2\ ft/s^2,\ y=-171\ ft\\ y=xtan\theta-\frac{x^2gsec^2\theta}{2u^2}\\ \therefore -171=-x(0.7536)-\frac{x^2g(1.568)}{2(225)}\\ \Rightarrow 0.1125x^2+0.7536x-171=0\\ x=35.78\ ft\ (can\ be\ calculated)\\ Horizontal\ range\ covered\ by\ the\ packet\ is\ 35.78\ ft.\\ So,\ the\ packet\ will\ fall\ 228-35.78=192\ ft\ short\ of\ his\ friend.$

40   A ball is projected from a point on the floor with a speed of $15$ $m/s$ at an angle of $60°$ with the horizontal. Will it hit a vertical wall $5$ $m$ away from the point of projection and perpendicular to the plane of projection without hitting the floor ? Will the answer differ if the wall is $22$ $m$ away ?

Solution :

Here $u=15\ \frac{m}{s},\ \theta=60^\circ,\ g=9.8\ \frac{m}{s^2}\\ Horizontal\ range\ X=\frac{u^2sin2\theta}{g}=\frac{15^2sin(2\times60^\circ)}{9.8}=19.88\ m.$

In first case the wall is $5\ m$ away from projection point, so it is in the horizontal range of projectile . So the ball will hit the wall. In second case (22 m away) wall is not within the horizontal range. So the ball would not hit the wall.

41   Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed $u$ at an angle $0$ with the horizontal.

Solution :

Total of flight T = $\frac{2usin\theta}{g}\\$ Average velocity = $\frac{change\ in\ displacement}{time}$ $\\$ from the figure , it can be said AB is horizontal . So there is no effect of vertical component of the velocity during this displacement. So because the body moves at a constant speed of $u'cos\theta ' $ in horizontal direction. The average velocity during this displacement will be $ucos\theta$ in the horizontal direction.

42   A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally ?

Solution :

During the motion of bomb its horizontal velocity $u$ remains constant and is same as that of aeroplane at every point of its path. Suppose the bomb explode i.e. reach the ground in time t. Distance travelled in horizontal direction by bomb $= ut = $ the distance travelled by aeroplane. so bomb explode vertically below the aeroplane. Suppose the aeroplane move making angle $\theta $ with horizontal. For both bomb and aeroplane , horizontal distance is u$cos\theta$t . t is time for bomb to reach the ground. so in this case also , the bomb will explode vertically below aeroplane.

43   A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of $1$ $m/s^2$ and the projection velocity in the vertical direction is $9.8$ $m/s$. How far behind the boy will the ball fall on the car ?

Solution :

Let the velocity of car be u when the all is thrown. Initial velocity of car is = Horizontal velocity of ball. Distance travelled by ball B $S_b=ut$ (in horizontal direction) And by car $S_c = ut +\frac{1}{2}at^2$ where $t\rightarrow time\ of\ flight\ of\ ball\ in\ air.\\ \therefore$ Car has travelled extra distance $S_c-S_b=\frac{1}{2}at^2.\\ $ Ball can be considered as a projectile having $\theta = 90^\circ$. $\therefore t=\frac{2usin\theta}{g}=\frac{2\times9.8}{9.8}=2\ sec.\\ \therefore S_c-S_b=\frac{1}{2}at^2=2\ m.\\$ $\therefore$ The ball will drop $2m$ behind the body.

44   A staircase contains three steps each $10$ $cm$ high and $20$ $cm$ wide (figure 3-E9). What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane ?

Solution :

At minimum velocity it will move just touching point E reaching the ground. $A$ is origin of reference coordinate. it $u$ is the minimum speed. $X=40,\ Y=-20,\ \theta=0^\circ\\ \therefore Y=xtan\theta-g\frac{x^2sec^2\theta}{2u^2}\ (because\ g=10\ \frac{m}{s^2}=1000\ \frac{cm}{s^2})\\ \Rightarrow -20=xtan\theta-\frac{1000\times40^2\times1}{2u^2}\\ \Rightarrow u=200\ \frac{cm}{s}=2\ m/s\\ \therefore The minimum horizontal velocity is 2\ m/s.$

45   A person is standing on a truck moving with a constant velocity of $14.7$ $m/s$ on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved $58.8$ $m$. Find the speed and the angle of projection $\\$(a) as seen from the truck, $\\$(b) as seen from the road

Solution :

a) As seen from the truck the ball moves vertically upwards comes back. Time taken = time taken by truck to cover $58.8\ m$. $\therefore time= \frac{s}{v}=\frac{58.8}{14.7}=4\ sec.\ (V=14.7\ m/s\ of\ truck)\\ u=?,\ v=0,\ g=-9.8\ m/s^2\ (goind\ upward),\ t=\frac{4}{2}=2\ sec.\\ v=u+at\Rightarrow 0=u-9.8\times2\Rightarrow u=19.6\ m/s./ (vertical\ upward\ velocity).$

Taking vertical component of velocity into consideration. $\\$ $y=\frac{0^2-(19.6)^2}{2\times(-9.8)}= 19.6\ m\ [from\ (a)]\\ \therefore y=usin\theta t-\frac{1}{2}gt^2\\ \Rightarrow 19.6=usin\theta(2)-\frac{1}{2}(9.8)2^2\Rightarrow2usin\theta=19.6\times2\\ \Rightarrow usin\theta=19.6 ...(ii)\\ \frac{usin\theta}{ucos\theta}=tan\theta\Rightarrow\frac{19.6}{14.7}=1.333\\ \Rightarrow \theta=tan^{-1}(1.333)=53^\circ\\$ Again $ucos\theta=14.7\\ \Rightarrow u=\frac{14.7}{ucos53^\circ}=24.42\ m/s.\\$ The speed of ball is $24.42\ m/s$ at an angle $53^\circ$ with horizontal as seen from the road.

b) From road it seems to be projectile motion . Total time of flight = $4\ sec$ In this time horizontal range covered $58.8\ m=x \\ \therefore X=ucos\theta t\\ \Rightarrow ucos\theta=14.7 ..(1)\\$

46   The benches of a gallery in a cricket stadium are $1\ m$ wide and $1\ m$ high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at $35\ m/s$ at an angle of $53°$ with the horizontal. The benches are perpendicular to the plane of motion and the first bench is $110\ m$ from the batsman. On which bench will the ball hit ?

Solution :

From the equation , y can be calculated.$\\$ $\therefore y =5\\ \Rightarrow n-1=5\Rightarrow n=6.\\$ The ball will drop in sixth bench.

$\theta=53^\circ,\ so cos53^\circ = \frac{3}{5}\\$ $Sec^{2}\theta=\frac{25}{9}\ and\ tan\theta=\frac{4}{3}\\$ Suppose the ball lands on nth bench $\\$ So, $y=(n-1)1\ ...(1)\ $[ball starting point $1\ m$ above ground] $\\$ Again $y=xtan\theta-\frac{gx^2sec^2\theta}{2u^2}\ [x=110+n-1=110+y] \\ \Rightarrow y=(110+y)(\frac{4}{3})-\frac{10(110+y)^2(\frac{25}{9})}{2\times35^2}\\ \Rightarrow=\frac{440}{3}+\frac{4}{3}y-\frac{250(110+y)^2}{18\times35^2}\\$

47   A man is sitting on the shore of a river. He is in the line of a $1.0$ $m$ long boat and is $5.5$ $m$ away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of $10$ $m/s$, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Solution :

When the apple just touches the end B of the boat. $\\$ $ x=5\ m,\ u=10\ m/s,\ g=10\ m/s^2,\ \theta=?\\ x=\frac{u^2sin2\theta}{g}\\ \Rightarrow 5=\frac{10^2sin2\theta}{10}\Rightarrow 5=10\ sin\ 2\theta\\ \Rightarrow sin2\theta=\frac{1}{2}\Rightarrow sin\ 30^\circ\ or\ sin150^\circ\\ \Rightarrow \theta=15^\circ\ or\ 75^\circ\\$ Similarly for end C, $x=6\ m$ $\\$ $Then\ 2\theta_1=sin^{-1}(\frac{gx}{u^2})=sin^{-1}(0.6)=182^\circ\ or\ 71^\circ .\\$ So, for a successful shot, $\theta$ may vary from $15^\circ\ to\ 18^\circ\ or\ 71^\circ\ to\ 75^\circ\ .$

48   A river $400$ $m$ wide is flowing at a rate of $2.0$ $m/s$. A boat is sailing at a velocity of $10$ $m/s$ with respect to the water, in a direction perpendicular to the river. $\\$(a) Find the time taken by the boat to reach the opposite bank.$\\$ (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank ?

Solution :

b) The boat will reach at point C. $\\ In\ \Delta ABC,\ tan\ \theta=\frac{BC}{AB}=\frac{BC}{400}=\frac{1}{5}\\ \Rightarrow BC=\frac{400}{5}=80\ m.$

a) Here the boat moves with the resultant velocity R. But the vertical component $10\ m/s$ takes him to the opposite shore. $\\$ $tan\ \theta=\frac{2}{10}=\frac{1}{5}\\ Velocity\ = 10\ m/s\\ distance=400\ m\\ Time=\frac{400}{10}=40\ sec.\\$

49   A swimmer wishes to cross a $500$ $m$ wide river flowing at $5$ $km/h$. His speed with respect to water is $3\ km/h.$ $\\$ (a) If he heads in a direction making an angle 0 with the flow, find the time he takes to cross the river. $\\$(b) Find the shortest possible time to cross the river

Solution :

b) Here vertical component of velocity i.e. $3\ km/hr$ takes him to opposite side. $\\$ $TIme=\frac{Distance}{Velocity}=\frac{0.5}{3}=0.16\ hr\\ \therefore 0.16\ hr=60\times0.16=9.6=10\ minute.$

a) The vertical component $3sin\theta $ takes him to opposite side. $\\$ $Distance =0.5\ km,\ Velocity =3sin\theta\ km/h\\ Time=\frac{Distance}{Velocity}=\frac{0.5}{3sin\theta}hr\\ =\frac{10}{sin\theta}\ min.$

50   Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Solution :

Velocity of man $\vec V_{m}=3\ km/hr$ $\\$ BD horizontal distance for resultant velocity R.$\\$ X-component of resultant $R_x=5+3\ cos\theta\\ t=\frac{0.5}{3sin\theta}\\$ which is same for horizontal component of velocity. $\\$ $H=BD=(5+3cos\theta)(\frac{0.5}{3\ sin\theta})=\frac{5+3cos\theta}{6sin\theta}\\$ $For\ H\ to\ be\ min\ (\frac{dH}{d\theta})=0\\ \Rightarrow \frac{d}{d\theta}(\frac{5+3cos\theta}{6sin\theta})=0\\$ $\Rightarrow -18(sin^2\theta+cos^2\theta)-30\ cos\theta=0\\$ $\Rightarrow -30cos\theta=18 \Rightarrow cos\theta=-\frac{18}{30}=\ -\frac{3}{5}\\ Sin\ theta=\sqrt{1-cos^2\theta}=\frac{4}{5}\\$ $\therefore H=\frac{5+3cos\theta}{6sin\theta}=\frac{5+3(-\frac{3}{5})}{6\times(\frac{4}{5})}=\frac{2}{3}\ km.$

51   An aeroplane has to go from a point $A$ to another point $B,\ 500\ km$ away due $30°$ east of north. A wind is blowing due north at a speed of $20\ m/s$. The air-speed of the plane is $150\ m/s$. $\\$(a) Find the direction in which the pilot should head the plane to reach the point B. $\\$(b) Find the time taken by the plane to go from $A$ to $B$.

Solution :

$\therefore \frac{20}{sin\ A}= \frac{150}{sin\ 30^\circ} \Rightarrow sinA= \frac{20}{150}sin30^\circ= \frac{20}{150}\times \frac{1}{2}= \frac{1}{15}\\ \Rightarrow A=sin^{-1}(\frac{1}{15})$

b) $sin^{-1}(\frac{1}{15})=3^\circ 48'\\ \Rightarrow 30^\circ+ 3^\circ 48'=33^\circ 48'\\ R=\sqrt{150^2+20^2+2(150)20cos33^\circ 48'}=167\ m/s.\\ Time =\frac{s}{v}=\frac{500000}{167}=2994\ sec=49=50\ min.$

In resultant direction $\vec R$ the plane rech the point $B$.$\\$ $Velocity\ of\ wind\ \vec V_w=20\ m/s\\$ $Velocity\ of\ aeroplane\ \vec V_a=150\ m/s\\ In\ \Delta ACD\ according\ to\ sine\ formula$

a) The direction is $sin^{-1}(\frac{1}{15})$ east of the line $AB$ $\\$