Motion in a Straight Line — NEET Physics PYQ MCQs with Solutions

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 $\dfrac{1}{v_1}+\dfrac{1}{v_2}=\dfrac{2}{v}$.

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 $t=\dfrac{t_1 t_2}{t_1+t_2}$.

Solution: Velocity $v=At+Bt^2=\dfrac{dx}{dt}$, so $dx=(At+Bt^2)\,dt$. Integrating from 1 to 2, $\Delta x=A\left[\dfrac{t^2}{2}\right]_1^2+B\left[\dfrac{t^3}{3}\right]_1^2=\dfrac{A}{2}(4-1)+\dfrac{B}{3}(8-1)=\dfrac{3A}{2}+\dfrac{7B}{3}$.

Solution: Velocity of each car: $V_P=\dfrac{dX_P}{dt}=a+2bt$ and $V_Q=\dfrac{dX_Q}{dt}=f-2t$. Setting $V_P=V_Q$: $a+2bt=f-2t\Rightarrow t=\dfrac{f-a}{2(b+1)}$.

Solution: Given $v=\beta x^{-2n}$. $a=\dfrac{dv}{dt}=\dfrac{dx}{dt}\cdot\dfrac{dv}{dx}=v\dfrac{dv}{dx}=(\beta x^{-2n})(-2n\beta x^{-2n-1})=-2n\beta^2 x^{-4n-1}$.

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$^2$.

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$^2$.

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, $a\propto v^{3/2}$.

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$^{-1}$.

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, $v_x=f_0 T-\dfrac{f_0}{T}\cdot\dfrac{T^2}{2}=f_0 T-\dfrac{f_0 T}{2}=\dfrac{1}{2}f_0 T$.

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 $=2XY$, $v_{av}=\dfrac{2XY}{XY\left(\frac{v_u+v_d}{v_u v_d}\right)}=\dfrac{2v_u v_d}{v_u+v_d}$.

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.

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.

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.

Solution: Velocity $v=\dfrac{ds}{dt}=9t^2+14t+14$. Acceleration $a=\dfrac{dv}{dt}=18t+14$. At $t=1$ s, $a=18(1)+14=32\,\text{m/s}^2$.

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, $2a-6bt=0\Rightarrow t=\dfrac{2a}{6b}=\dfrac{a}{3b}$.

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 $v_{max}=t\left(\dfrac{\alpha\beta}{\alpha+\beta}\right)$.

Solution: Velocity $v=\dfrac{ds}{dt}=3t^2-12t+3$. Acceleration $a=\dfrac{dv}{dt}=6t-12$. For $a=0$, $t=2$ s. Then $v=3(2)^2-12(2)+3=-9\,\text{ms}^{-1}$.

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.

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.

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.

Solution: For two equal halves, $v_{av}=\dfrac{2v_1v_2}{v_1+v_2}=\dfrac{2\times40\times60}{40+60}=48$ km/h.

Solution: Given, initial velocity of car $u=0$, acceleration $a=5$ m/s$^2$. At $t=4$ s, $v=u+at=0+(5)(4)=20$ m/s, so the car's velocity when the ball is dropped is $20$ m/s. The ball's velocity in the $x$-direction $v_x=20$ m/s (due to the car). In the $y$-direction the acceleration is $g=10$ m/s$^2$, so $v_y=u+a_y t=0+10\times2=20$ m/s (over the $2$ s from $t=4$ s to $t=6$ s). Net velocity $v=\sqrt{v_x^2+v_y^2}=\sqrt{20^2+20^2}=20\sqrt{2}$ m/s, and the acceleration is $g=10$ m/s$^2$.

Solution: Initial velocity $u=0$. Distance covered in the $n$th second, $s_n=u+\dfrac{a}{2}(2n-1)=\dfrac{a}{2}(2n-1)$. Distance covered in the $(n+1)$th second, $s_{n+1}=u+\dfrac{a}{2}[2(n+1)-1]=\dfrac{a}{2}(2n+1)$. Dividing, $\dfrac{s_n}{s_{n+1}}=\dfrac{2n-1}{2n+1}$.

Solution: Time to cross the window $t=0.1$ s, $h=1.5$ m. If $u$ is the velocity at the topmost point of the window, then $h=ut+\dfrac{1}{2}gt^2$ $\Rightarrow 1.5=u\times0.1+\dfrac{1}{2}\times10\times(0.1)^2$ $\Rightarrow 1.5=0.1u+0.05$ $\Rightarrow u=\dfrac{1.5-0.05}{0.1}=\dfrac{1.45}{0.1}=14.5$ m/s.

Solution: Given $u=20$ m/s, $v=80$ m/s. From $v^2=u^2+2gh$, $h=\dfrac{v^2-u^2}{2g}=\dfrac{(80)^2-(20)^2}{2\times10}=300$ m.

Solution: Let $h$ be the height through which the coin is dropped. From $h=ut+\dfrac{1}{2}gt^2$ with $u=0$, $t=\sqrt{\dfrac{2h}{g}}$, so $t\propto\dfrac{1}{\sqrt{g}}$. As the elevator moves uniformly its velocity is constant, so its acceleration is zero and the relative acceleration $g'=g\pm0=g$. Hence the time to reach the floor is the same, i.e. $t_1=t_2$.

Solution: For $0<t<1$ s the velocity increases from $0$ to $6$ ms$^{-1}$. After the field is reversed, for $1<t<2$ s the velocity decreases from $6$ to $0$, and for $2<t<3$ s it increases to $-6$ ms$^{-1}$. Magnitude of acceleration $|a|=\left|\dfrac{v-u}{t}\right|=\dfrac{6-0}{1}=6$ ms$^{-2}$. Using $s=ut+\dfrac{1}{2}at^2$: $s_1=0+\dfrac{1}{2}\times6\times1^2=3$ m; $s_2=6\times1-\dfrac{1}{2}\times6\times1^2=3$ m; $s_3=0-\dfrac{1}{2}\times6\times1^2=-3$ m. Net displacement $=3+3-3=3$ m, so average velocity $=\dfrac{3}{3}=1$ ms$^{-1}$. Total distance $=9$ m, so average speed $=\dfrac{9}{3}=3$ ms$^{-1}$.

Solution: For free fall $u=0$. Distance in first $5$ s, $h_1=\dfrac{1}{2}g(5)^2=\dfrac{25}{2}g$. Distance in first $10$ s, $s_2=\dfrac{1}{2}g(10)^2=\dfrac{100}{2}g$, so distance in the second $5$ s, $h_2=s_2-h_1=\dfrac{100}{2}g-\dfrac{25}{2}g=\dfrac{75}{2}g$. Distance in first $15$ s, $s_3=\dfrac{1}{2}g(15)^2=\dfrac{225}{2}g$, so distance in the last $5$ s, $h_3=s_3-s_2=\dfrac{225}{2}g-\dfrac{100}{2}g=\dfrac{125}{2}g$. Therefore $h_1:h_2:h_3=\dfrac{25}{2}g:\dfrac{75}{2}g:\dfrac{125}{2}g=1:3:5$, i.e. $h_1=\dfrac{h_2}{3}=\dfrac{h_3}{5}$.

Solution: Given $g=10$ m/s$^2$ and $h=20$ m. $v=\sqrt{2gh}=\sqrt{2\times10\times20}=\sqrt{400}=20$ m/s.