J J Science Academy

1) A particle starts moving along a straight line with velocity ' $u$ ' and a constant acceleration ' $a$ '. After time ' $t$ ', it acquires velocity ' $v$ '. Then the average velocity in time $t$ is

1) $\overline{v}=\frac{u+v}{2}$

2) $\overline{v}=\frac{u-v}{2}$

3) $\overline{v}=\frac{u^{2}+v}{2}$

4) $\overline{v}=\frac{u^{2}-v}{2}$

$$ v^2=u^2+2as \Rightarrow s=\frac{v^2-u^2}{2a} $$ $$ v=u+at \Rightarrow t=\frac{v-u}{a} $$ $$ \overline{v}=\frac{s}{t}=\frac{(v+u)(v-u)}{2a} \times \frac{a}{v-u}=\frac{v+u}{2} $$

2) 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

1) $\left(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\right)t$

2) $\left(\frac{\alpha^{2}-\beta^{2}}{\alpha \beta}\right)t$

3) $\frac{(\alpha+\beta) t}{\alpha \beta}$

4) $\frac{\alpha \beta t}{\alpha+\beta}$

Let the car accelerate at rate $\alpha$ for time $t_1$ then maximum velocity attained, $$ v=0+\alpha t_1=\alpha t_1 $$ Now, the car decelerates at a rate $\beta$ for time $(t-t_1)$ and finally comes to rest. Then, $$ 0=v-\beta\left(t-t_1\right) $$ $$ \therefore 0=\alpha t_1-\beta t+\beta t_1 \\ \therefore t_1=\frac{\beta}{\alpha+\beta} t \\ \therefore v=\frac{\alpha \beta}{\alpha+\beta} t $$

3) A body moving along negative X -axis, with speed $u$ , is subjected to unit acceleration along positive X -axis, when it is at the origin. The time it takes to return to origin is

1) $\frac{u}{2}$

2) $u$

3) $\frac{3 u}{2}$

4) $2u$

$$ T=\frac{u-(-u)}{a}=\frac{2 u}{a} $$ As $a=1 m / s^2$, $$ T=2u $$

4) A body moving along negative X -axis, with speed $u$ , is subjected to unit acceleration along positive $X$-axis, when it is at the origin, the distance it travels before reaching origin is

1) $u^2$

2) $2 u^2$

3) $\frac{u^{2}}{2}$

4) $4 u^2$

$$ v^2=u^2-2as=0 $$ $$ \Rightarrow s \propto v^2 $$ $$ \frac{s_1}{s_2} = (\frac{v_1}{v_2})^2 = 3^2 = 9 $$

5) A particle experiences, constant acceleration for 20 second after starting from rest. If it travels a distance $S_1$ in the first 10 second and a distance $S_2$ the next 10 second, then

1) $S_2=S_1$

2) $S_2=2 S_1$

3) $S_2=3 S_1$

4) $S_2=4 S_1$

The distance travelled in first 10 second, $$ S_1=\frac{1}{2}g(10)^2=50g $$ The distance travelled in first 20 second, $$ S=\frac{1}{2}g(20)^2=200g $$ $$ S_2=S-S_1=200g-50g=150g=3 S_1 $$

6) Tripling the speed of a motor car, multiplies the distance needed for stopping it by

1) 3

2) 6

3) 9

4) some other number

$$ v^2=u^2-2as=0 \Rightarrow s \propto v^2 $$ $$ \frac{s_1}{s_2} = (\frac{v_1}{v_2})^2 = 3^2 = 9 $$

7) A body is accelerated at constant rate, from rest. It travels a distance of 20 m in 5 sec . The distance it will travel in next 5 sec will be

1) 60 m

2) 80 m

3) 120 m

4) 20 m

s2 = 3s = 3 x 20= 60 m

8) A bullet fired into a fixed target loses one-third of its velocity after penetrating 6 cm . How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

1) 1.5 cm

2) 0.75 cm

3) 3 cm

4) 4 cm

First case: $$ \left(\frac{v}{3}\right)^2=v^2-(2a \times 6) $$ $$ 12a=\frac{8 v^2}{9} $$ $$ v^2=\frac{12 \times 9}{8}a=\frac{27}{2}a $$ Second case: $$ 0=\left(\frac{v}{3}\right)^2-2ax $$ $$ v^2=18ax $$ From (i) and (ii) $$ 18ax=\frac{27}{2}a \\ x=\frac{3}{4}=0.75 cm $$

9) A conveyor belt is moving horizontally at a speed of $4 m/s$ . If a box of mass 30 kg is gently laid on it, it takes 0.1 second for the box to come to rest on the belt. The distance moved by the box on the conveyor belt (in m ) is

1) 0

2) 0.2

3) 0.4

4) 0.8

$$ v=u+at $$ $$ 0=-4+a \times (0.1) $$ $$ a=40 m / s^2 $$ $$ s=\frac{v^2}{2a}=\frac{16}{80}=0.2 m $$

10) A bus, starts from rest, with a constant acceleration $2 m/s^2$ and travels for 10 s . The distance travelled by it in last 3 seconds of its motion is

1) 36 m

2) 51 m

3) 49 m

4) 59 m

$$ s_{10}=\frac{1}{2}(2)(10)^2=100 m $$ $$ s_7=\frac{1}{2}(2)(7)^2=49 m $$ Distance travelled in last 3 sec is $$ s_{10}-s_7=100-49=51 m $$

11) A body starts from rest and has an acceleration $20 cm/s^2$ . What is the distance covered by the body in first 8 s ?

1) 160 cm

2) 640 cm

3) 1280 cm

4) 1640 cm

$$ s=ut+\frac{1}{2}at^2 $$ $$ =0+\left(\frac{1}{2} \times 20 \times(8)^2\right) \\ =640 cm $$

12) Two cars A and B at rest at same point initially. If A starts with uniform velocity of $40 m/sec$ and B starts in the same direction with constant acceleration of $4 m/s^2$ , then B will catch A after how much time

1) 10 s

2) 20 s

3) 30 s

4) 35 s

Let A and B will meet after time t sec. it means the distance travelled by both will be equal. $$ s_A=ut=40t $$ $$ s_B=\frac{1}{2}at^2=\frac{1}{2} \times 4 \times t^2=2t^2 $$ As $s_A=s_B$ $$ 40t = 2t^2 $$ $$ t=20 s $$

13) At a certain a particle has a speed of $18 m/s$ in positive x -direction and 2.4 s later its speed was $30 m/s$ in the opposite direction. What is the magnitude of the average acceleration of the particle during the 2.4 s interval?

1) 20 m/s^2

2) 10 m/s^2

3) 5 m/s^2

4) 2.5 m/s^2

$$ v=u+at $$ $$ -30=18+(2.4 t) $$ $$ t=\frac{-30-18}{2.4}=-\frac{48}{2.4}=-20 m / s^2 $$

14) A particle experiences, constant acceleration for 20 second after starting from rest. If it travels a distance $S_1$ in the first 10 second and a distance $S_2$ the next 10 second, then

1) $S_2=S_1$

2) $S_2=2 S_1$

3) $S_2=3 S_1$

4) $S_2=4 S_1$

The distance travelled in first 10 second, $S_1=\frac{1}{2}g(10)^2=50g$ The distance travelled in first 20 second, $S=\frac{1}{2}g(20)^2=200g$ $$ S_2=S-S_1=200g-50g=150g=3 S_1 $$

15) Tripling the speed of a motor car, multiplies the distance needed for stopping it by

1) 3

2) 6

3) 9

4) some other number

$$ v^2=u^2-2as=0 \Rightarrow s \propto v^2 $$ $$ \frac{s_1}{s_2}=\left(\frac{v_1}{v_2}\right)^2=3^2=9 $$

16) A stone travels a distance of 25 m in the last second of its motion. It falls from a height of

1) 55 m

2) 60 m

3) 20 m

4) 45 m

The distance travelled in $n^{th}$ second, $s_n=\frac{1}{2}g(2n-1)$ $$ 25=5(2n-1) $$ $$ 2n-1=5 $$ $$ n=3 $$ $$ s=\frac{1}{2}gt^2=\frac{1}{2} \times 10 \times 9=45 m $$

17) An uniformly accelerated body travels a distance of 24 m in $5^{th}$ second of its motion and 44 m in 15 $^{th}$ sec of its motion. The acceleration of the particle will be

1) 1 m/s^2

2) 2 m/s^2

3) 4 m/s^2

4) 0.5 m/s^2

$$ s_n=u+\frac{a}{2}(2n-1) $$ $$ 24=u+\frac{9a}{2} $$ $$ 44=u+\frac{29a}{2} $$ (ii) - (i) gives $$ 20=10a $$ $$ a=2 m / s^2 $$

18) A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the $5^{th}$ second to that covered in 5 seconds is

1) $\frac{9}{25}$

2) $\frac{3}{25}$

3) $\frac{25}{9}$

4) $\frac{1}{25}$

s = = 1 a (2n - 1) = 9a $$ S5=\frac{1}{2}at^2=\frac{25a}{2} $$ $$ s_5 $$

19) Velocity-time curves plotted for two bodies are straight lines passing through origin. The slope of v-t curve for two bodies X and Y is 3 and 2 respectively. In a given time interval, the ratio of displacement of X to that of Y is

1) $\frac{9}{4}$

2) $\frac{4}{9}$

3) 2:3

4) 3:2

Slope of v-t curve = acceleration $$ s \propto a $$ $$ s_1/s_2=a_1/a_2=3/2 $$

20) A ball rolls from the top of a stair way with a horizontal velocity $u m/s$ . If the steps are h m high and b m wide, the ball will hit the edge of the $n^{th}$ step, if

1) $n=\frac{2 h u}{g b^{2}}$

2) $n=\frac{2 h u^{2}}{g b^{2}}$

3) $n=\frac{2 h u^{2}}{g b}$

4) $n=\frac{h u^{2}}{g b^{2}}$

$$ nh=\frac{1}{2}gt^2 \quad nb=ut $$ From these two equation we get, $$ nh=\frac{g}{2}\left(\frac{nb}{u}\right)^2 \\ n=\frac{2hu^2}{gb^2} $$

21) A particle start moving from rest state along a straight line under the action of a constant force and travel distance $x$ in first 5 seconds. The distance travelled by it in next five seconds will be

1) x

2) 2x

3) 3x

4) 4x

$$ s=ut+\frac{1}{2}at^2 $$ $$ s_5=0+\frac{1}{2}a(5)^2 $$ $$ x=\frac{25a}{2} $$ $$ s_{10}=0+\frac{1}{2}a(10)^2=\frac{100a}{2}=4x $$ $$ \Delta s=s_{10}-s_5=4x-x=3x $$

22) A car, moving with a speed of $50 km/hr$ , can be stopped by brakes after at least 6 m . If the same car is moving at a speed of $100 km/hr$ , the minimum stopping distance is

1) 6 m

2) 12 m

3) 18 m

4) 24 m

$$ s \propto u^2 $$ $$ \frac{s_1}{s_2}=\left(\frac{u_1}{u_2}\right)^2=\left(\frac{50}{100}\right)^2=\frac{1}{4} $$ $$ s_2=4 s_1=4 \times 6=24 m $$

23) A bullet fired into a fixed target loses one-third of its velocity after penetrating 6 cm . How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

1) 1.5 cm

2) 0.75 cm

3) 3 cm

4) 4 cm

First case: $$ \left(\frac{v}{3}\right)^2=v^2-(2a \times 6) $$ $$ 12a=\frac{8 v^2}{9} $$ $$ v^2=\frac{12 \times 9}{8}a=\frac{27}{2}a $$ Second case: $$ 0=\left(\frac{v}{3}\right)^2-2ax $$ $$ v^2=18ax $$ From (i) and (ii) $$ 18ax=\frac{27}{2}a \\ x=\frac{3}{4}=0.75 cm $$

24) A body moving with a speed of $10 m/s$ , long -Z - axis, experiences an acceleration of $5 \hat{i}-3 \hat{j}+4 \hat{k} m / s^2$ . Its displacement at the end of 2 s , is

1) $5 \hat{i}-3 \hat{j}+6 \hat{k}$

2) $10 \hat{i}-6 \hat{j}+12 \hat{k}$

3) $10 \hat{i}-3 \hat{j}+12 \hat{k}$

4) $10 \hat{i}-6 \hat{j}-12 \hat{k}$

$$ \vec{u}=-10 \hat{k} $$ $$ \vec{s}=\vec{u} t+\frac{1}{2} \vec{a} t^2=[(-10 \hat{k}) \times 2]+\frac{1}{2} \times(5 \hat{i}-3 \hat{j}+4 \hat{k}) \times(2)^2 $$ $$ =(-20 \hat{k})+(10 \hat{i}-6 \hat{j}+8 \hat{k})=10 \hat{i}-6 \hat{j}-12 \hat{k} $$

25) A uniformly accelerated car, starting from rest, covers a distance S , in 4 sec. The distance it travels in next 4 sec, is

1) S

2) 2 S

3) 3 S

4) 4 S

Distance travelled in successive interval has a ratio of 1:3:5 .....

26) A bus, starts from rest, with a constant acceleration $2 m/s^2$ and travels for 10 s . The distance travelled by it in last 3 seconds of its motion is

1) 36 m

2) 51 m

3) 49 m

4) 59 m

$$ s_{10}=\frac{1}{2} \times(2) \times(10)^2=100 m $$ $$ s_7=\frac{1}{2} \times(2) \times(7)^2=49 m $$ Distance travelled in last $3 sec=s_{10}-s_7=100-49=51 m$

27) A ball is dropped from a bridge 125 meter above a river. After the ball has been falling for two second, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time? ( $g=10 m/s^2$ )

1) 39 m/s

2) 55.5 m/s

3) 26.7 m/s

4) 19.8 m/s

For first ball, $$ s=\frac{1}{2}gt^2 $$ $$ 125=\frac{1}{2} \times 10 \times t^2 $$ $$ t=5 s $$ For second ball, $t_1=3 s$ $$ s_1=ut_1+\frac{1}{2}gt_1^2 $$ $$ 125=3u+\frac{1}{2} \times 10 \times 9 $$ $$ u=\frac{80}{3}=26.7 m / s $$

28) A body is moving with uniform acceleration describes 40 m in the first 5 sec and 65 m in next 5 sec . Its initial velocity will be

1) 4 m/s

2) 2.5 m/s

3) 5.5 m/s

4) 11 m/s

$$ a=\frac{s_2-s_1}{t^2}=\frac{65-40}{(5)^2}=1 m / s^2 $$ From equation (i), we get, $$ s_1=ut+\frac{1}{2}at^2 $$ $$ 40=5u+\left(\frac{1}{2} \times 1 \times 25\right) $$ $$ 5u=40-12.5=27.5 $$ $$ \therefore u=5.5 m / s $$

29) The initial velocity of a body moving along a straight line is $7 m/s$ . It has a uniform acceleration of $4 m/s^2$ . The distance covered by the body in the $5^{th}$ second of its motion is

1) 25 m

2) 35 m

3) 50 m

4) 85 m

$$ s_n=u+\frac{a}{2}[2n-1] = 7+\frac{4}{2}[(2\times5)-1]=7+18=25 m $$