Kinematics Equations of Uniformly Accelerated Motion — NEET Physics PYQ MCQs with Solutions

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.

Solution: For the first ball $u=0$, so $s_1=\dfrac{1}{2}gt_1^2=\dfrac{1}{2}g(18)^2$. For the second ball, initial velocity $=v$ and $t_2=18-6=12$ s, so $s_2=vt_2+\dfrac{1}{2}gt_2^2=12v+\dfrac{1}{2}g(12)^2$. Since they meet, $s_1=s_2$: $\dfrac{1}{2}g(18)^2=12v+\dfrac{1}{2}g(12)^2$ $\Rightarrow v=74$ ms$^{-1}$.

Solution: Since the body starts from rest $u=0$, so $s=\dfrac{1}{2}at^2$. Then $s_1=\dfrac{1}{2}a(10)^2$ ...(i) and $s_2=\dfrac{1}{2}a(20)^2$ ...(ii). Dividing (i) by (ii): $\dfrac{s_1}{s_2}=\dfrac{(10)^2}{(20)^2}$, so $s_2=4s_1$.

Solution: Using $v^2-u^2=2as$: $(20)^2-(10)^2=2\times a\times135$ $\Rightarrow a=\dfrac{300}{270}=\dfrac{10}{9}$ ms$^{-2}$. Using $v-u=at$: $20-10=\dfrac{10}{9}\times t$ $\Rightarrow t=9$ s.

Solution: Distance travelled in the $n$th second, $s_n=u+\dfrac{1}{2}a(2n-1)$. Here $u=0$, $a=\dfrac{4}{3}$. So $s_3=0+\dfrac{1}{2}\times\dfrac{4}{3}\times(2\times3-1)=\dfrac{10}{3}$ m.

Solution: For free fall from a height, $u=0$. From the second equation of motion $h=\dfrac{1}{2}gt^2$, so $\dfrac{h_1}{h_2}=\left(\dfrac{t_1}{t_2}\right)^2$. Given $h_1=16$ m, $h_2=25$ m, so $\dfrac{t_1}{t_2}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$.

Solution: At maximum height the final speed is zero, so from $v=u-gt$ with $v=0$, $t=\dfrac{u}{g}$. In $2$ s, $u=g\times2=9.8\times2=19.6$ m/s. With a throw speed of $19.6$ m/s, when the $3$rd ball is thrown the $1$st just reaches the ground, so only $2$ balls are in the air. To keep more than $2$ balls in the air the ball must be thrown with a speed greater than $19.6$ m/s.

Solution: Let the ball take $T$ seconds to reach maximum height $H$. From $v=u-gT$ with $v=0$ at $H$, $T=\dfrac{u}{g}$. Velocity attained after $(T-t)$ s is $v'=u-g(T-t)=u-gT+gt=u-u+gt=gt$. Hence the distance travelled in the last $t$ sec of ascent, $CB=v't-\dfrac{1}{2}gt^2=(gt)t-\dfrac{1}{2}gt^2=gt^2-\dfrac{1}{2}gt^2=\dfrac{1}{2}gt^2$.

Solution: Let $u$ be the initial velocity and $H$ the maximum height. At $h=\dfrac{H}{2}$, $v_1=10$ m/s. From $v_1^2=u^2-2gh$: $(10)^2=u^2-2g\dfrac{H}{2}$ ...(i). At height $H$, $v_2=0$: $0=u^2-2gH$ ...(ii). Subtracting (ii) from (i): $(10)^2=2g\dfrac{H}{2}$, so $H=\dfrac{(10)^2}{g}=\dfrac{100}{10}=10$ m.

Solution: By conservation of energy, kinetic energy = work done in stopping: $\dfrac{1}{2}mv^2=Fs$. For the same retarding force $s\propto v^2$, so $\dfrac{s_2}{s_1}=\left(\dfrac{v_2}{v_1}\right)^2=\left(\dfrac{80}{40}\right)^2=4$. Thus $s_2=4s_1=4\times2=8$ m.

Solution: Given $u=0$, $t=20$ s, $v=144$ km/h $=40$ m/s. From $v=u+at$: $a=\dfrac{v-u}{t}=\dfrac{40-0}{20}=2$ m/s$^2$. Then $s=ut+\dfrac{1}{2}at^2=0+\dfrac{1}{2}\times2\times(20)^2=400$ m.

Solution: Initial velocity $u=40$ m/s, acceleration $a=-g=-10$ m/s$^2$, time $=2$ s. From $v=u+at$: $v=40-10\times2=20$ m/s.

Solution: When an object falls freely under gravity, its speed depends only on the height of fall and is independent of mass. As all objects fall through the same height, by conservation of mechanical energy $\dfrac{1}{2}mv^2=mgl$, so $v=\sqrt{2gl}$ and $v_1:v_2:v_3=1:1:1$.

Solution: Let $t$ be the time interval of two drops. For third drop to fall, $5=\dfrac{1}{2}g(2t)^2$ [as $u=0$], or $\dfrac{1}{2}gt^2=\dfrac{5}{4}$ ...(i). Let $x$ be the distance through which the second drop falls for time $t$, then $x=\dfrac{1}{2}gt^2=\dfrac{5}{4}$ m [from Eq.(i)]. Thus, height of second drop from ground $=5-\dfrac{5}{4}=\dfrac{15}{4}=3.75$ m.

Solution: Time taken by the body to reach the ground from some height is the same as taken to reach that height. Hence, time to reach the ground from its maximum height is $5$ s.

Solution: Initial velocity of stone $u=0$. Time to reach the ground $t=4$ s. Acceleration $a=+g=10$ m/s$^2$ (as motion of body is along the acceleration due to gravity). $\therefore$ Height of tower $h=ut+\dfrac{1}{2}gt^2=(0\times4)+\dfrac{1}{2}\times10\times4^2=80$ m.

Solution: Distance travelled by the body in $n$th second is given by $s_n=u+\dfrac{a}{2}(2n-1)$. Here, $u=0$. $\therefore$ For 4th s, $s_4=\dfrac{a}{2}(2\times4-1)$ and for 3rd s, $s_3=\dfrac{a}{2}(2\times3-1)$. Hence, $\dfrac{s_4}{s_3}=\dfrac{(2\times4-1)}{(2\times3-1)}=\dfrac{7}{5}$.

Solution: Let the body fall through the height of the tower in $t$ seconds. From $s_n=u+\dfrac{a}{2}(2n-1)$, total distance travelled in the last $2$ s of fall is $s=s_t+s_{(t-1)}=\left[0+\dfrac{g}{2}(2t-1)\right]+\left[0+\dfrac{g}{2}(2(t-1)-1)\right]=\dfrac{g}{2}(2t-1)+\dfrac{g}{2}(2t-3)=\dfrac{g}{2}(4t-4)=\dfrac{10}{2}\times4(t-1)$. or $40=20(t-1)$ or $t=2+1=3$ s. Distance travelled in $t$ sec is $s=ut+\dfrac{1}{2}at^2=0+\dfrac{1}{2}\times10\times3^2=45$ m.

Solution: Distance travelled in $n$th sec is given by $s_n=u+\dfrac{1}{2}a(2n-1)$. Here, $u=0$, acceleration due to gravity $a=9.8$ m/s$^2$. $\therefore$ For 4th s, $s_4=\dfrac{1}{2}\times9.8(2\times4-1)$ and for 5th s, $s_5=\dfrac{1}{2}\times9.8(2\times5-1)$. $\therefore \dfrac{s_4}{s_5}=\dfrac{7}{9}$.

Solution: Let $x$ be the total distance between points $P$ and $Q$ and $v$ be the velocity of car while passing a certain middle point $R$ of $PQ$. If $a$ is the acceleration of the car, then for part $PQ$, $40^2-30^2=2ax$ or $a=\dfrac{350}{x}$ ...(i). For part $RQ$, $40^2-v^2=\dfrac{2ax}{2}$ ...(ii). Putting value of $a$ from Eq.(i) in Eq.(ii), we have $40^2-v^2=2\left(\dfrac{350}{x}\right)\dfrac{x}{2}$ or $40^2-v^2=350$ or $v^2=1250$ $\Rightarrow v=25\sqrt{2}$ km/h.