Motion in a Straight Line

Class 11 NCERT Physics

NCERT

1   In which of the following examples of motion, can the body be considered approximately a point object:$\\$ (a) a railway carriage moving without jerks between two stations.$\\$ (b) a monkey sitting on top of a man cycling smoothly on a circular track.$\\$ (c) a spinning cricket ball that turns sharply on hitting the ground.$\\$ (d) a tumbling beaker that has slipped off the edge of a table.

Solution :

(a) The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.$\\$ (b) The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.$\\$ (c) The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.$\\$ (d) The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.

2   The position-time $(x-t)$ graphs for two children $A$ and $B$ returning from their school $O$ to their homes $P$ and $Q$ respectively are shown in figure. Choose the correct entries in the brackets below;$\\$ (a) $(A/B)$ lives closer to the school than $(B/A)$$\\$ (b) $(A/B)$ starts from the school earlier than $(B/A)$$\\$ (c) $(A/B)$ walks faster than $(B/A)$$\\$ (d) $A$ and $B$ reach home at the (same/different) time$\\$ (e) $(A/B)$ overtakes $(B/A)$ on the road (once/twice).

Solution :

(a) As $OP < OQ, A$ lives closer to the school than $B.$$\\$ (b) For $x = 0, t = 0$ for $A$; while $t$ has some finite value for $B$. Therefore, $A$ starts from the school earlier than $B.$$\\$ (c) Since the velocity is equal to slope of $x-t$ graph in case of uniform motion and slope of $x-t$ graph for $B$ is greater that that for $A$ =, hence $B$ walks faster than $A.$$\\$ (d) It is clear from the given graph that both $A$ and $B$ reach their respective homes at the same time.$\\$ (e) $B$ moves later than $A$ and his/her speed is greater than that of $A.$ From the graph, it is clear that $B$ overtakes$A$ only once on the road.

3   A woman starts from her home at $9.00$ am, walks with a speed of $5 km /h ^{- 1}$ on a straight road up to her office $2.5 km$ away, stays at the office up to $5.00$ pm, and returns home by an auto with a speed of $25 km /h^{- 1 }$. Choose suitable scales and plot the $x-t$ graph of her motion

Solution :

Speed of the woman $= 5 km/h$$\\$ Distance between her office and home $= 2.5 km$$\\$ Time taken = Distance / Speed$\\$ $= \frac{2.5} {5 }= 0.5 h = 30 min$$\\$ It is given that she covers the same distance in the evening by an auto. Now, speed of the auto $= 25 km/h$$\\$ Time taken = Distance /Speed $\\$ $= \frac{2.5 }{ 25} = \frac{1}{ 10 }= 0.1 h = 6 min$$\\$ The suitable $x-t$ graph of the motion of the woman is shown in the given figure.

4   A drunkard walking in a narrow lane takes $5$ steps forward and $3$ steps backward, followed again by $5$ steps forward and $3$ steps backward, and so on. Each step is $1$ m long and requires $1$ s. Plot the $x-t$ graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit $13 m$ away from the start.

Solution :

Distance covered with $1$ step = $1 m$$\\$ Time taken = $1 s$$\\$ Time taken to move first $5 m$ forward = $5 s$$\\$ Time taken to move $3 m$ backward = $3 s$$\\$ Net distance covered =$ 5 - 3 = 2 m$$\\$ Net time taken to cover $2 m = 8 s$$\\$ Drunkard covers $2 m$ in $8 s.$$\\$ Drunkard covered $4 m$ in $16 s.$$\\$ Drunkard covered $6 m$ in $24 s.$$\\$ Drunkard covered $8 m$ in $32 s.$$\\$ In the next $5 s$, the drunkard will cover a distance of $5 m$ and a total distance of $13 m$ and falls into the pit.$\\$ Net time taken by the drunkard to cover $13 m = 32 + 5 = 37 s$$\\$ The $x-t$ graph of the drunkard’s motion can be shown as:

5   A jet airplane travelling at the speed of $500 km /h^{ -1}$ ejects its products of combustion at the speed of $1500 km /h^{-1}$ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?

Solution :

Speed of the jet airplane, $= 500 km/h$$\\$ Relative speed of its products of combustion with respect to the plane,$\\$ Speed of its products of combustion with respect to the ground$\\$ Relative speed of its products of combustion with respect to the airplane,$\\$ The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.