Terms Related to Motion — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Terms Related to Motion MCQs with step-by-step solutions (22 questions). Part of Motion in a Straight Line. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Terms Related to Motion · medium
A person travelling in a straight line moves with a constant velocity $v_1$ for certain distance '$x$' and with a constant velocity $v_2$ for next equal distance. The average velocity $v$ is given by the relation
A. $\dfrac{1}{v}=\dfrac{1}{v_1}+\dfrac{1}{v_2}$
B. $\dfrac{2}{v}=\dfrac{1}{v_1}+\dfrac{1}{v_2}$ ✓ Correct
C. $\dfrac{v}{2}=\dfrac{v_1+v_2}{2}$
D. $v=\sqrt{v_1 v_2}$
Solution: For distance $x$ the person moves with $v_1$ and for another $x$ with $v_2$. Total distance $D=x+x=2x$, total time $T=t_1+t_2=\dfrac{x}{v_1}+\dfrac{x}{v_2}$. Average velocity $v=\dfrac{2x}{\frac{x}{v_1}+\frac{x}{v_2}}=\dfrac{2}{\frac{1}{v_1}+\frac{1}{v_2}}$, hence $\dfrac{1}{v_1}+\dfrac{1}{v_2}=\dfrac{2}{v}$.
Q2 — Terms Related to Motion · medium
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_1$. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_2$. The time taken by her to walk up on the moving escalator will be
A. $\dfrac{t_1+t_2}{2}$
B. $\dfrac{t_1 t_2}{t_2-t_1}$
C. $\dfrac{t_1 t_2}{t_2+t_1}$ ✓ Correct
D. $t_1-t_2$
Solution: Speed of walking $=\dfrac{h}{t_1}=v_1$ and speed of escalator $=\dfrac{h}{t_2}=v_2$. When she walks over the running escalator, $t=\dfrac{h}{v_1+v_2}$, so $\dfrac{1}{t}=\dfrac{v_1+v_2}{h}=\dfrac{1}{t_1}+\dfrac{1}{t_2}$, giving $t=\dfrac{t_1 t_2}{t_1+t_2}$.
Q3 — Terms Related to Motion · medium
If the velocity of a particle is $v=At+Bt^2$, where $A$ and $B$ are constants, then the distance travelled by it between 1s and 2s is
A. $3A+7B$
B. $\dfrac{3}{2}A+\dfrac{7}{3}B$ ✓ Correct
C. $\dfrac{A}{2}+\dfrac{B}{3}$
D. $\dfrac{3}{2}A+4B$
Solution: Velocity $v=At+Bt^2=\dfrac{dx}{dt}$, so $dx=(At+Bt^2)\,dt$. Integrating from 1 to 2, $\Delta x=A\left[\dfrac{t^2}{2}\right]_1^2+B\left[\dfrac{t^3}{3}\right]_1^2=\dfrac{A}{2}(4-1)+\dfrac{B}{3}(8-1)=\dfrac{3A}{2}+\dfrac{7B}{3}$.
Q4 — Terms Related to Motion · medium
Two cars $P$ and $Q$ start from a point at the same time in a straight line and their positions are represented by $X_P(t)=at+bt^2$ and $X_Q(t)=ft-t^2$. At what time do the cars have the same velocity?
A. $\dfrac{a-f}{1+b}$
B. $\dfrac{a+f}{2(b-1)}$
C. $\dfrac{a+f}{2(1+b)}$
D. $\dfrac{f-a}{2(1+b)}$ ✓ Correct
Solution: Velocity of each car: $V_P=\dfrac{dX_P}{dt}=a+2bt$ and $V_Q=\dfrac{dX_Q}{dt}=f-2t$. Setting $V_P=V_Q$: $a+2bt=f-2t\Rightarrow t=\dfrac{f-a}{2(b+1)}$.
Q5 — Terms Related to Motion · medium
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v(x)=\beta x^{-2n}$ where, $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$, is given by
A. $-2n\beta^2 x^{-2n-1}$
B. $-2n\beta^2 x^{-4n-1}$ ✓ Correct
C. $-2\beta^2 x^{-2n+1}$
D. $-2n\beta^2 e^{-4n+1}$
Solution: Given $v=\beta x^{-2n}$. $a=\dfrac{dv}{dt}=\dfrac{dx}{dt}\cdot\dfrac{dv}{dx}=v\dfrac{dv}{dx}=(\beta x^{-2n})(-2n\beta x^{-2n-1})=-2n\beta^2 x^{-4n-1}$.
Q6 — Terms Related to Motion · medium
The motion of a particle along a straight line is described by equation $x=8+12t-t^3$ where, $x$ is in metre and $t$ in sec. The retardation of the particle when its velocity becomes zero, is
A. $24$ ms$^{-2}$
B. zero
C. $6$ ms$^{-2}$
D. $12$ ms$^{-2}$ ✓ Correct
Solution: $v=\dfrac{dx}{dt}=12-3t^2$ and $a=\dfrac{dv}{dt}=-6t$. Velocity is zero when $12-3t^2=0\Rightarrow t=2$ s. At $t=2$ s, $a=-6\times2=-12$ m/s$^2$, so retardation $=12$ m/s$^2$.
Q7 — Terms Related to Motion · medium
A body is moving with velocity 30 m/s towards East. After 10s, its velocity becomes 40 m/s towards North. The average acceleration of the body is
A. $7$ m/s$^2$
B. $\sqrt{7}$ m/s$^2$
C. $5$ m/s$^2$ ✓ Correct
D. $1$ m/s$^2$
Solution: Average acceleration $a=\dfrac{|\vec{v}_f-\vec{v}_i|}{\Delta t}=\dfrac{\sqrt{30^2+40^2}}{10}=\dfrac{\sqrt{900+1600}}{10}=\dfrac{\sqrt{2500}}{10}=5$ m/s$^2$.
Q8 — Terms Related to Motion · medium
A particle moves a distance $x$ in time $t$ according to the equation $x=(t+5)^{-1}$. The acceleration of particle is proportional to
A. $(\text{velocity})^{3/2}$ ✓ Correct
B. $(\text{distance})^2$
C. $(\text{distance})^{-2}$
D. $(\text{velocity})^{2/3}$
Solution: Given $x=(t+5)^{-1}$. Differentiating, $\dfrac{dx}{dt}=v=-(t+5)^{-2}$. Again, $\dfrac{d^2x}{dt^2}=a=2(t+5)^{-3}$. Comparing the two, $a\propto v^{3/2}$.
Q9 — Terms Related to Motion · medium
A bus is moving with a speed of $10$ ms$^{-1}$ on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what speed should the scooterist chase the bus?
A. $20$ ms$^{-1}$ ✓ Correct
B. $40$ ms$^{-1}$
C. $25$ ms$^{-1}$
D. $10$ ms$^{-1}$
Solution: Let $v$ be the relative velocity of scooter ($s$) w.r.t. bus ($B$): $v=v_s-v_B$. Relative velocity $=\dfrac{\text{displacement}}{\text{time}}=\dfrac{1000}{100}=10$ ms$^{-1}$. Hence $v_s=v+v_B=10+10=20$ ms$^{-1}$.
Q10 — Terms Related to Motion · medium
A particle moving along $x$-axis has acceleration $f$, at time $t$, given by $f=f_0\left(1-\dfrac{t}{T}\right)$, where $f_0$ and $T$ are constants. The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$, the particle's velocity ($v_x$) is
A. $f_0 T$
B. $\dfrac{1}{2}f_0 T^2$
C. $f_0 T^2$
D. $\dfrac{1}{2}f_0 T$ ✓ Correct
Solution: $\dfrac{dv}{dt}=f_0\left(1-\dfrac{t}{T}\right)$. Integrating, $v=f_0 t-\dfrac{f_0}{T}\cdot\dfrac{t^2}{2}+c$. At $t=0$, $v=0\Rightarrow c=0$. $f=0$ gives $t=T$. Substituting, $v_x=f_0 T-\dfrac{f_0}{T}\cdot\dfrac{T^2}{2}=f_0 T-\dfrac{f_0 T}{2}=\dfrac{1}{2}f_0 T$.
Q11 — Terms Related to Motion · medium
A car moves from $X$ to $Y$ with a uniform speed $v_u$ and returns to $X$ with a uniform speed $v_d$. The average speed for this round trip is
A. $\dfrac{2v_d v_u}{v_d+v_u}$ ✓ Correct
B. $\sqrt{v_u v_d}$
C. $\dfrac{v_d v_u}{v_d+v_u}$
D. $\dfrac{v_u+v_d}{2}$
Solution: Average speed $=\dfrac{\text{total distance}}{\text{time taken}}$. With $t_1+t_2=\dfrac{XY}{v_u}+\dfrac{XY}{v_d}=XY\left(\dfrac{v_u+v_d}{v_u v_d}\right)$ and total distance $=2XY$, $v_{av}=\dfrac{2XY}{XY\left(\frac{v_u+v_d}{v_u v_d}\right)}=\dfrac{2v_u v_d}{v_u+v_d}$.
Q12 — Terms Related to Motion · medium
The position $x$ of a particle w.r.t. time $t$ along $x$-axis is given by $x=9t^2-t^3$, where $x$ is in metre and $t$ in sec. What will be the position of this particle when it achieves maximum speed along the $+x$ direction?
A. 32 m
B. 54 m ✓ Correct
C. 81 m
D. 24 m
Solution: Given $x=9t^2-t^3$. Differentiating w.r.t. time, speed $v=\dfrac{dx}{dt}=18t-3t^2$. Again differentiating, acceleration $a=\dfrac{dv}{dt}=18-6t$. Speed is maximum when acceleration is zero, so $18-6t=0\Rightarrow t=3$ s. Putting this in the position equation, $x=9(3)^2-(3)^3=81-27=54$ m.
Q13 — Terms Related to Motion · medium
A particle moves along a straight line $OX$. At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x=40+12t-t^3$. How long would the particle travel before coming to rest?
A. 24 m
B. 40 m
C. 56 m ✓ Correct
D. 16 m
Solution: Velocity $v=\dfrac{dx}{dt}=12-3t^2$. The particle comes to rest when $v=0$, so $12-3t^2=0\Rightarrow t^2=4\Rightarrow t=2$ s. Distance travelled before coming to rest, $x=40+12(2)-(2)^3=40+24-8=56$ m.
Q14 — Terms Related to Motion · medium
The displacement $x$ of a particle varies with time $t$ as $x=ae^{-\alpha t}+be^{\beta t}$, where $a,b,\alpha$ and $\beta$ are positive constants. The velocity of the particle will
A. decrease with time
B. be independent of $\alpha$ and $\beta$
C. drop to zero when $\alpha=\beta$
D. increase with time ✓ Correct
Solution: Velocity $v=\dfrac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}=A+B$, where $A=-a\alpha e^{-\alpha t}$ and $B=b\beta e^{\beta t}$. The term $A$ decreases while the term $B$ increases with time, so the velocity goes on increasing with time.
Q15 — Terms Related to Motion · medium
A particle moves along a straight line such that its displacement at any time $t$ is given by $s=3t^3+7t^2+14t+5$. The acceleration of the particle at $t=1$ s is
A. $18\,\text{m/s}^2$
B. $32\,\text{m/s}^2$ ✓ Correct
C. $29\,\text{m/s}^2$
D. $24\,\text{m/s}^2$
Solution: Velocity $v=\dfrac{ds}{dt}=9t^2+14t+14$. Acceleration $a=\dfrac{dv}{dt}=18t+14$. At $t=1$ s, $a=18(1)+14=32\,\text{m/s}^2$.
Q16 — Terms Related to Motion · medium
The position $x$ of a particle varies with time $t$, as $x=at^2-bt^3$. The acceleration of the particle will be zero at time $t$ equals to
A. zero
B. $\dfrac{a}{3b}$ ✓ Correct
C. $\dfrac{2a}{3b}$
D. $\dfrac{a}{b}$
Solution: $x=at^2-bt^3$, so velocity $v=\dfrac{dx}{dt}=2at-3bt^2$ and acceleration $a=\dfrac{dv}{dt}=2a-6bt$. Setting acceleration to zero, $2a-6bt=0\Rightarrow t=\dfrac{2a}{6b}=\dfrac{a}{3b}$.
Q17 — Terms Related to Motion · medium
A car accelerates from rest at a constant rate $\alpha$ for some time, after which it decelerates at a constant rate $\beta$ and comes to rest. If the total time elapsed is $t$, then the maximum velocity acquired by the car is
A. $\left(\dfrac{\alpha^2+\beta^2}{\alpha\beta}\right)t$
B. $\left(\dfrac{\alpha^2-\beta^2}{\alpha\beta}\right)t$
C. $\dfrac{(\alpha+\beta)t}{\alpha\beta}$
D. $\dfrac{\alpha\beta t}{\alpha+\beta}$ ✓ Correct
Solution: On a $v$-$t$ graph, $OA$ is the accelerated part and $AB$ the decelerated part. At the peak, $v_{max}=\alpha t_1=\beta t_2$. Since $t=t_1+t_2=\dfrac{v_{max}}{\alpha}+\dfrac{v_{max}}{\beta}=v_{max}\left(\dfrac{\alpha+\beta}{\alpha\beta}\right)$, we get $v_{max}=t\left(\dfrac{\alpha\beta}{\alpha+\beta}\right)$.
Q18 — Terms Related to Motion · medium
A particle moves along a straight line such that its displacement at any time $t$ is given by $s=(t^3-6t^2+3t+4)$ m. The velocity when the acceleration is zero, is
A. $3\,\text{ms}^{-1}$
B. $-12\,\text{ms}^{-1}$
C. $42\,\text{ms}^{-1}$
D. $-9\,\text{ms}^{-1}$ ✓ Correct
Solution: Velocity $v=\dfrac{ds}{dt}=3t^2-12t+3$. Acceleration $a=\dfrac{dv}{dt}=6t-12$. For $a=0$, $t=2$ s. Then $v=3(2)^2-12(2)+3=-9\,\text{ms}^{-1}$.
Q19 — Terms Related to Motion · medium
A train of 150 m length is going towards North direction at a speed of 10 m/s. A parrot flies at the speed of 5 m/s towards South direction parallel to the railways track. The time taken by the parrot to cross the train is
A. 12 s
B. 8 s
C. 15 s
D. 10 s ✓ Correct
Solution: Relative velocity of the parrot w.r.t. the train $=v_A-v_B=[10-(-5)]=15\,\text{ms}^{-1}$. Time taken to cross the train $=\dfrac{150}{15}=10$ s.
Q20 — Terms Related to Motion · medium
A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and the last one-third at 60 km/h. The average speed of the bus is
A. 9 km/h
B. 16 km/h
C. 18 km/h ✓ Correct
D. 48 km/h
Solution: Times for the three equal thirds are $t_1=\dfrac{s/3}{10}$, $t_2=\dfrac{s/3}{20}$, $t_3=\dfrac{s/3}{60}$. Average speed $=\dfrac{s}{t_1+t_2+t_3}=\dfrac{s}{s/18}=18$ km/h.
Q21 — Terms Related to Motion · medium
A car moves a distance of 200 m. It covers the first-half of the distance at speed 40 km/h and the second-half of distance at speed $v$ km/h. The average speed is 48 km/h. Find the value of $v$.
A. 56 km/h
B. 60 km/h ✓ Correct
C. 50 km/h
D. 48 km/h
Solution: With $t_1=\dfrac{100}{40}$ and $t_2=\dfrac{100}{v}$, the average speed $48=\dfrac{200}{\frac{100}{40}+\frac{100}{v}}$. This gives $\dfrac{1}{40}+\dfrac{1}{v}=\dfrac{1}{24}\Rightarrow\dfrac{1}{v}=\dfrac{1}{60}\Rightarrow v=60$ km/h.
Q22 — Terms Related to Motion · medium
A car covers the first-half of the distance between two places at 40 km/h and other half at 60 km/h. The average speed of the car is
A. 40 km/h
B. 48 km/h ✓ Correct
C. 50 km/h
D. 60 km/h
Solution: For two equal halves, $v_{av}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2\times40\times60}{40+60}=48$ km/h.