Chapter3 Motion In A Straight Line
Chapter 3
MOTION IN A
STRAIGHT LINE
MCQ I
3.1 Among the four graphs (Fig. 3.1), there is only one graph for which average velocity over the time intervel $(O, T)$ can vanish for a suitably chosen $T$. Which one is it?
(a)
(c)
(b)
(d)
Fig. 3.1
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Answer: (b)3.2 A lift is coming from $8^{\text {th }}$ floor and is just about to reach $4^{\text {th }}$ floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct?
(a) $x<0, v<0, a>0$
(b) $x>0, v<0, a<0$
(c) $x>0, v<0, a>0$
(d) $x>0, v>0, a<0$
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Answer: (a)3.3 In one dimensional motion, instantaneous speed $v$ satisfies $0 \leq v<v_{0}$.
(a) The displacement in time $T$ must always take non-negative values.
(b) The displacement $x$ in time $T$ satisfies $-v_{\mathrm{o}} \mathrm{T}<x<v_{\mathrm{o}} \mathrm{T}$.
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.
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Answer: (b)3.4 A vehicle travels half the distance $L$ with speed $V_{1}$ and the other half with speed $V_{2}$, then its average speed is
(a) $\frac{V_{1}+V_{2}}{2}$
(b) $\frac{2 V_{1}+V_{2}}{V_{1}+V_{2}}$
(c) $\frac{2 V_{1} V_{2}}{V_{1}+V_{2}}$
(d) $\frac{L\left(V_{1}+V_{2}\right)}{V_{1} V_{2}}$
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Answer: (c)3.5 The displacement of a particle is given by $x=(t-2)^{2}$ where $x$ is in metres and $t$ in seconds. The distance covered by the particle in first 4 seconds is
(a) $4 \mathrm{~m}$
(b) $8 \mathrm{~m}$
(c) $12 \mathrm{~m}$
(d) $16 \mathrm{~m}$
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Answer: (b)3.6 At a metro station, a girl walks up a stationary escalator in time $t_{1}$. If she remains stationary on the escalator, then the escalator take her up in time $t_{2}$. The time taken by her to walk up on the moving escalator will be
(a) $\left(t_{1}+t_{2}\right) / 2$
(b) $t_{1} t_{2} /\left(t_{2}-t_{1}\right)$
(c) $t_{1} t_{2} /\left(t_{2}+t_{1}\right)$
(d) $t_{1}-t_{2}$
MCQ II
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Answer: (c)3.7 The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.
(a) Quantity B may represent time.
(b) Quantity A is velocity if motion is uniform.
(c) Quantity A is displacement if motion is uniform.
(d) Quantity A is velocity if motion is uniformly accelerated.
Fig. 3.2
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Answer: (a), (c), (d)3.8 A graph of $x$ versus $t$ is shown in Fig. 3.3. Choose correct alternatives from below.
(a) The particle was released from rest at $t=0$.
(b) At B, the acceleration $a>0$.
(c) At $\mathrm{C}$, the velocity and the acceleration vanish.
(d) Average velocity for the motion between A and D is positive.
(e) The speed at $\mathrm{D}$ exceeds that at $\mathrm{E}$.
Fig. 3.3
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Answer: (a), (c), (e)3.9 For the one-dimensional motion, described by $x=t-\sin t$
(a) $x(t)>0$ for all $t>0$.
(b) $v(\mathrm{t})>0$ for all $t>0$.
(c) $a(t)>0$ for all $t>0$.
(d) $v(\mathrm{t})$ lies between 0 and 2 .
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Answer: (a), (d)3.10 A spring with one end attached to a mass and the other to a rigid support is stretched and released.
(a) Magnitude of acceleration, when just released is maximum.
(b) Magnitude of acceleration, when at equilibrium position, is maximum.
(c) Speed is maximum when mass is at equilibrium position.
(d) Magnitude of displacement is always maximum whenever speed is minimum.
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Answer: (a), (c)3.11 A ball is bouncing elastically with a speed $1 \mathrm{~m} / \mathrm{s}$ between walls of a railway compartment of size $10 \mathrm{~m}$ in a direction perpendicular to walls. The train is moving at a constant velocity of $10 \mathrm{~m} / \mathrm{s}$ parallel to the direction of motion of the ball. As seen from the ground,
(a) the direction of motion of the ball changes every 10 seconds.
(b) speed of ball changes every 10 seconds.
(c) average speed of ball over any 20 second interval is fixed.
(d) the acceleration of ball is the same as from the train.
VSA
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Answer: (b), (c), (d)3.12 Refer to the graphs in Fig 3.1. Match the following.
Graph Characteristic
(a) (i) has $v>0$ and $a<0$ throughout.
(b) (ii) has $x>0$ throughout and has a point with $v=0$ and a point with $a=0$.
(c) (iii) has a point with zero displacement for $t>0$.
(d) (iv) has $v<0$ and $a>0$.
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Answer: (a) (iii), (b) (ii), (c) iv, (d) (i)3.13 A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with time. (Take acceleration in the backward direction as positive).
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Answer:
3.14 Give examples of a one-dimensional motion where
(a) the particle moving along positive $x$-direction comes to rest periodically and moves forward.
(b) the particle moving along positive $x$-direction comes to rest periodically and moves backward.
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Answer: (i) $x(t)=t-\sin t$
(ii) $x(t)=\sin t$
3.15 Give example of a motion where $x>0, v<0, a>0$ at a particular instant.
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Answer: $x(t)=\mathrm{A}+B e^{-\gamma t} ; \mathrm{A}>\mathrm{B}, \gamma>0$ are suitably chosen positive constants.3.16 An object falling through a fluid is observed to have acceleration given by $a=g-b v$ where $g=$ gravitational acceleration and $\mathrm{b}$ is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?
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Answer: $v=g / \mathrm{b}$3.17 A ball is dropped and its displacement vs time graph is as shown Fig. 3.4 (displacement $x$ is from ground and all quantities are +ve upwards).
(a) Plot qualitatively velocity vs time graph.
(b) Plot qualitatively acceleration vs time graph.
Fig. 3.4
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Answer: The ball is released and is falling under gravity. Acceleration is $-g$, except for the short time intervals in which the ball collides with
ground, and when the impulsive force acts and produces a large acceleration.
3.18 A particle executes the motion described by $x(t)=x_{o}\left(1-e^{-\gamma t}\right) ; t \geq 0$, $x_{0}>0$.
(a) Where does the particle start and with what velocity?
(b) Find maximum and minimum values of $x(t), v(t), a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.
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Answer: (a) $\mathrm{x}=0, v=\gamma x_{o}$3.19 A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has a speed of 18 $\mathrm{m} / \mathrm{h}$ while the other has the speed of $27 \mathrm{~km} / \mathrm{h}$. The bird starts moving from first car towards the other and is moving with the speed of $36 \mathrm{~km} / \mathrm{h}$ and when the two cars were separted by $36 \mathrm{~km}$. What is the total distance covered by the bird? What is the total displacement of the bird?
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Answer: Relative speed of cars $=45 \mathrm{~km} / \mathrm{h}$, time required to meet $=\frac{36 \mathrm{~km}}{45 \mathrm{~km} / \mathrm{h}}=0.80 \mathrm{~h}$
Thus, distance covered by the bird $=36 \mathrm{~km} / \mathrm{h} \times 0.8 \mathrm{~h}=28.8 \mathrm{~km}$.
3.20 A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 $\mathrm{m} / \mathrm{s}$, the (horizontal) distance between the two buildings is $10 \mathrm{~m}$ and the height difference is $9 \mathrm{~m}$, will he be able to land on the next building? (take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )
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Answer: Suppose that the fall of $9 \mathrm{~m}$ will take time $t$. Hence
$y-y_{o}=v_{o y}-\frac{g t^{2}}{2}$
Since $v_{o y}=0$,
$t=\sqrt{\frac{2\left(y-y_{o}\right)}{g}} \rightarrow \sqrt{\frac{2 \times 9 \mathrm{~m}}{10 \mathrm{~m} / \mathrm{s}^{2}}}=\sqrt{1.8} \approx 1.34$ seconds.
In this time, the distance moved horizontally is
$x-x_{o}=v_{o x} t=9 \mathrm{~m} / \mathrm{s} \times 1.34 \mathrm{~s}=12.06 \mathrm{~m}$.
Yes-he will land.
3.21 A ball is dropped from a building of height $45 \mathrm{~m}$. Simultaneously another ball is thrown up with a speed $40 \mathrm{~m} / \mathrm{s}$. Calculate the relative speed of the balls as a function of time.
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Answer: Both are free falling. Hence, there is no acceleration of one w.r.t. another. Therefore, relative speed remains constant $(=40 \mathrm{~m} / \mathrm{s})$.3.22 The velocity-displacement graph of a particle is shown in Fig. 3.5.
(a) Write the relation between $v$ and $x$.
(b) Obtain the relation between acceleration and displacement and plot it.
Fig. 3.5
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Answer: $v=\left(-v_{0} / x_{0}\right) x+v_{0}, a=\left(v_{0} / x_{0}\right)^{2} x-v_{0}^{2} / x_{0}$
The variation of $a$ with $x$ is shown in the figure. It is a straight line with a positive slope and a negative intercept.
3.23 It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
(a) If a rain drop falls from such a height freely under gravity, what will be its speed? Also calculate in $\mathrm{km} / \mathrm{h} .\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
(b) A typical rain drop is about $4 \mathrm{~mm}$ diameter. Momentum is mass $\mathrm{x}$ speed in magnitude. Estimate its momentum when it hits ground.
(c) Estimate the time required to flatten the drop.
(d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
(e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is $5 \mathrm{~cm}$.
(Assume that umbrella is circular and has a diameter of $1 \mathrm{~m}$ and cloth is not pierced through !!)
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Answer: (a) $v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 1000}=141 \mathrm{~m} / \mathrm{s}=510 \mathrm{~km} / \mathrm{h}$.
(b) $\quad m=\frac{4 \pi}{3} r^{3} \rho=\frac{4 \pi}{3}\left(2 \times 10^{-3}\right)^{3}\left(10^{3}\right)=3.4 \times 10^{-5} \mathrm{~kg}$.
$$ P=m v \approx 4.7 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \approx 5 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s} . $$
(c) $\quad$ Diameter $\approx 4 \mathrm{~mm}$
$$ \Delta t \approx d / v=28 \mu \mathrm{s} \approx 30 \mu \mathrm{s} $$
(d) $F=\frac{\Delta P}{\Delta t}=\frac{4.7 \times 10^{-3}}{28 \times 10^{-6}} \approx 168 \mathrm{~N} \approx 1.7 \times 10^{2} \mathrm{~N}$.
(e) Area of cross-section $=\pi d^{2} / 4 \approx 0.8 \mathrm{~m}^{2}$.
With average separation of $5 \mathrm{~cm}$, no. of drops that will fall almost simultaneously is $\frac{0.8 \mathrm{~m}^{2}}{\left(5 \times 10^{-2}\right)^{2}} \approx 320$.
Net force $\approx 54000$ N (Practically drops are damped by air viscosity).
3.24 A motor car moving at a speed of $72 \mathrm{~km} / \mathrm{h}$ can not come to a stop in less than $3.0 \mathrm{~s}$ while for a truck this time interval is $5.0 \mathrm{~s}$. On a higway the car is behind the truck both moving at $72 \mathrm{~km} / \mathrm{h}$. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is $0.5 \mathrm{~s}$.
(Comment: This is to illustrate why vehicles carry the message on the rear side. “Keep safe Distance”)
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Answer: Car behind the truck
Regardation of truck $=\frac{20}{5}=4 \mathrm{~ms}^{-2}$
Regardation of car $=\frac{20}{3} \mathrm{~ms}^{-2}$
Let the truck be at a distance $x$ from the car when breaks are applied
Distance of truck from A at $t>0.5 \mathrm{~s}$ is $\quad x+20 t-2 t^{2}$.
Distance of car from A is $\quad 10+20(t-0.5)-\frac{10}{3}(t-0.5)^{2}$.
If the two meet
$x+20 t-2 t^{2}=10+20 t-10-\frac{10}{3} t^{2}+\frac{10}{3} t-0.25 \times \frac{10}{3}$.
$x=-\frac{4}{3} t^{2}+\frac{10}{3} t-\frac{5}{6}$.
To find $x_{\text {min }}$,
$\frac{d x}{d t}=-\frac{8}{3} t+\frac{10}{3}=0$
which gives $t_{\min }=\frac{10}{8}=\frac{5}{4} \mathrm{~s}$.
Therefore, $x_{\min }=-\frac{4}{3}\left(\frac{5}{4}\right)^{2}+\frac{10}{3} \times \frac{5}{4}-\frac{5}{6}=\frac{5}{4}$.
Therefore, $x>1.25 \mathrm{~m}$.
Second method: This method does not require the use of calculus.
If the car is behind the truck,
$V_{\text {car }}=20-(20 / 3)(t-0.5)$ for $t>0.5 \mathrm{~s}$ as car declerate only after $0.5 \mathrm{~s}$.
$V_{\text {truck }}=20-4 t$
Find $t$ from equating the two or from velocity vs time graph. This yields $t=5 / 4 \mathrm{~s}$.
In this time truck would travel truck,
$S_{\text {truck }}=20(5 / 4)-(1 / 2)(4)(5 / 4)^{2}=21.875 m$ and car would travel, $S_{\text {car }}=20(0.5)+20(5 / 4-0.5)-$ $\left(\frac{1}{2}\right)(20 / 3) \times\left(\frac{5}{4}-0.5\right)^{2}=23.125 \mathrm{~m}$
Thus $S_{\text {car }}-S_{\text {truck }}=1.25 \mathrm{~m}$.
If the car maintains this distance initially, its speed after $1.25 \mathrm{~s}$ will he always less than that of truck and hence collision never occurs.
3.25 A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time $t$ is given by $v(t)=2 t(3-t) ; 0<t<3$ and $v(\mathrm{t})=-(t-3)(6-t)$ for $3<\mathrm{t}<6 \mathrm{~s}$ in $\mathrm{m} / \mathrm{s}$. It repeats this cycle till it reaches the height of $20 \mathrm{~m}$.
(a) At what time is its velocity maximum?
(b) At what time is its average velocity maximum?
(c) At what time is its acceleration maximum in magnitude?
(d) How many cycles (counting fractions) are required to reach the top?
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Answer: (a) $(3 / 2) \mathrm{s}$, (b) $(9 / 4) \mathrm{s}$, (c) $0,3 \mathrm{~s}$, (d) 6 cycles.3.26 A man is standing on top of a building $100 \mathrm{~m}$ high. He throws two balls vertically, one at $t=0$ and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is $+15 \mathrm{~m}$ at $t=2 \mathrm{~s}$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
