102 + Interview Questions in GENERAL-KINEMATICS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. `1 : 1`B. `1 : 4`C. `1 : 8`D. `1 : 16`
Answer» Correct Answer - D Using `v^2 = u^2 + 2as`. If car is retarding and final velocity becomes zero. Then, the equation becomes `0 = u^2 - 2as rArr s = (u^2)/(2a)` or `s prop u^2` As `s prop u^2 rArr (s_1)/(s_2) = ((1)/(4))^2 = (1)/(16)`.

Discussion

2.	A body `A` starts from rest with an acceleration `a_1`. After `2` seconds, another body `B` starts from rest with an acceleration `a_2`. If they travel equal distances in the `5th` second, after the start of `A`, then the ratio `a_1 : a_2` is equal to :A. `5 : 9`B. `5 : 7`C. `9 : 5`D. `9 : 7`
Answer» Correct Answer - A According to problem Distance travelled by body `A` in `5^(th) sec` and distance travelled by body `B` in `3^(rd) sec` of its motion are equal. `0 + (a_1)/(2)(2 xx 5 - 1) = 0 + (a_2)/(2)[2 xx 3 - 1]` `9 a_1 = 5 a_2 rArr (a_1)/(a_2) = (5)/(9)`.

Discussion

3.	If a body starts from rest, the time in which it covers a particular displacement with uniform acceleration is :A. inversely alphaortional to the square root of the displacementB. inversely alphaortional to the dispalcementC. directly alphaortional to the displacementD. directly alphaortional to the square root of the displacement
Answer» Correct Answer - D Using `s = ut + (1)/(2) at^2` as initial velocity is zero hence, `s =(1)/(2) at^2 rArr t prop sqrt(s)`.

Discussion

4.	A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(2v(n + 1))/(n)`B. `(v(n + 1))/(n)`C. `(v(n - 1))/(n)`D. `(2 v(n - 1))/(n)`
Answer» Correct Answer - D `because v = 0 + na rArr a = (v)/(n)` Now, distance travelled in `n sec` `rArr S_n = (1)/(2) an^2` and distance travelled in `(n - 2) sec` `rArr S_(n - 2) = (1)/(2) a(n - 2)^2` `:.` Distance travelled in last two seconds =`S_n - S_(n - 2)` =`(1)/(2) an^2 - (1)/(2) a(n - 2)^2` =`(a)/(2)[n^2 - (n - 2)^2]` =`(a)/(2)[n + (n - 2)][n -(n - 2)]` =`a(2 n - 2)` =`(v)/(n)(2n - 2) = (2v(n - 1))/(n)`.

Discussion

5.	A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(v(n + 1))/(n)`B. `(2 v(n + 1))/(n)`C. `(2 v(n - 1))/(n)`D. `(v(n - 1))/(n)`
Answer» Correct Answer - C `v = 0 + na rArr a = (v)/(n)` displacement in last two seconds `s(n) - s(n - 2)` =`(0 + (1)/(2) an^2) - (0 + (1)/(2) a(n - 2)^2)` =`2a(n - 1) = 2v ((n - 1))/(n)`.

Discussion

6.	A body starting from rest moving with uniform acceleration has a displacement of `16 m` in first `4` seconds and `9 m` in first `3` seconds. The acceleration of the body is :A. `1 ms^-2`B. `2 ms^-2`C. `3 ms^-2`D. `4 ms^-2`
Answer» Correct Answer - B Distance traveled in `4th` second `= 16 - 9 = 7 m` Now `D_n = u + (a)/(2) (2n - 1)` Given `u = 0` `rArr 7 = 0 + (a)/(2) (2 xx 4 - 1) rArr a = 2 m//s^2`.

Discussion

7.	A body is moving from rest under constant acceleration and let `S_1` be the displacement in the first `(p - 1)` sec and `S_2` be the displacement in the first `p` sec. The displacement in `(p^2 - p + 1)` sec. will beA. `S_1 + S_2`B. `S_1 S_2`C. `S_1 - S_2`D. `S_1//S_2`
Answer» Correct Answer - A From `S = ut + (1)/(2) at^2` `S_1 = (1)/(2)a(P - 1)^2` and `S_2 = (1)/(2) a P^2" "["As" u = 0]` From `S_n = u + (a)/(2) (2n - 1)` `S_((P^2 - P + 1)^(th)) = (a)/(2)[2(P^2 - P + 1)-1]` =`(a)/(2) [2 P^2 - 2P + 1]` It is clear that `S_((P^2 - P + 1)^(th))= S_1 + S_2`.

Discussion

8.	A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`
Answer» Correct Answer - D `s = t^2 - 6t + 3t + 4` Velocity `v = (ds)/(dt) = 3t^2 - 6 xx 2t + 3 xx 1 + 0` =`3t^2 - 12t + 3` ….(i) Acceleration `a = (dv)/(dt) = 3 xx 2t - 12 xx 1 + 0 = 6t - 12` Acceleration is zero at time t given by `6t-12=0` `rArr t = 2 seconds` So, velocity v at `t = 2` seconds is `v = (3t^2 - 12t + 3)_(t = 2s)` =`3 xx (2)^2 - 12 xx 2 + 3 = -9 m//s`.

Discussion

9.	The displacement of a particle moving along x-axis is given by : `x = a + bt + ct^2` The acceleration of the particle is.A. bB. cC. `b + c`D. 2 c
Answer» Correct Answer - D `x = a + bt + ct^2` velocity `v = (dx)/(dt) = 0 + b + c. 2t = b + 2ct` acceleration `f = (dv)/(dt) = 0 + 2c = 2c`.

Discussion

10.	If the velocity of a particle is `v = At + Bt^2`, where `A` and `B` are constant, then the distance travelled by it between `1 s` and `2 s` is :A. `(3)/(2) A + 4B`B. `3 A + 7 B`C. `(3)/(2)A + (7)/(3) B`D. `(A)/(2) + (B)/(3)`
Answer» Correct Answer - C `V = At + Bt^2 rArr (dx)/(dt) = At + Bt^2` `rArr underset(0)overset(x)(int) dx = underset(1)overset(2)(int)(At + Bt^2) dt` `rArr x = (A)/(2) (2^2 - 1^2) + (B)/(3)(2^3 - 1^3) =(3 A)/(2) + (7 B)/(3)`.

Discussion

11.	The displacement of a particle is given by `y = a + bt + ct^2 - dt^4`. The initial velocity and acceleration are respectively.A. `b, -4d`B. `-b, 2c`C. `b,2c`D. `2c, -4d`
Answer» Correct Answer - C `y = a + bt + ct^2 - dt^4` `:. v = (dy)/(dt) = b+ 2ct - 4dt^3` and `a = (dv)/(dt) = 2c - 12 dt^2`. Hence, at `t = 0, v_("initial") = b and a_("initial") = 2 c`.

Discussion

12.	Refer `Q.139`. The average acceleration of the body from `t = 0` to `t = 15 s` is.A. `1.25 m//s^2`B. `4//7 m//s^2`C. `5//6 m//s^2`D. `7//6 m//s^2`
Answer» Correct Answer - D Area upto `15 s = 175 m//s = v_2` Velocity at `t = 15 s` `a_(av) = (v_2 - 0)/(15) = (17.5)/(15) = (7)/(6) m//s^2`.

Discussion

13.	Refer `Q.149`. At time `t_3`. Which car is moving faster ?A. car `A`B. car `B`C. same speedD. None of these
Answer» Correct Answer - B at `t_3`, slope of graph for car `B` is more than that of `A`. Hence car `B` is faster.

Discussion

14.	A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight areA. `(vec v)/(2) and (v)/(2)`B. `0 and (v)/(2)`C. `0 and 0`D. `(vec v)/(2) and 0`.
Answer» Correct Answer - B Average velocity = `0` because net displacement of the body is zero. Average speed `= ("Total distance covered")/("Time of flight")` =`(2 H_(max))/(2 u//g)` `rArr v_(av) = (2 u^2//2g)/(2u//g) rArr v_(av) = (u)/(2)` Velocity of projection = v (given) `:. v_(av) = (v)/(2)`.

Discussion

15.	A car moves from `X` to `Y` with a uniform speed `v_u` and returns to `Y` with a uniform speed `v_d`. The average speed for this round trip is :A. `(2 v_d v_u)/(v_d + v_u)`B. `sqrt(v_u v_d)`C. `(v_d v_u)/(v_d + v_u)`D. `(v_u + v_d)/(2)`
Answer» Correct Answer - A We define Average speed `= ("Distance travelled")/("Time taken") = (d)/(T)` Let `t_1` and `t_2` be times taken by the car to go from `X` to `Y` and then from `Y` to `X` respectively. Then, `t_1 + t_2 =[(XY)/(v_u)]+[(XY)/(v_d)] = XY((v_u + v_d)/(v_u v_d))` Total distance travelled `d=XY+XY=2XY` Therefore, average speed of the car for this round trip is. `v_(av) = (2 XY)/(XY((v_u + v_d)/(v_u v_d)))` or `v_(av) = (2 v_uv_d)/(v_u + v_d)`.

Discussion

16.	A ball is upward with such a velocity `v` that it returns to the thrower after after `3 s`. Take `g = 10 ms^-2`. Find the value of `v`.A. 15 m//sB. 20 m//sC. 10 m//sD. 5 m//s
Answer» Correct Answer - A Using `vec h = vec u t + (1)/(2) vec a t^2`, Since the displacement is zero, `0 = uT - (1)/(2) g t^2 rArr u = (gT)/(2)` `rArr u = (10 xx 3)/(2) = 15 m//s`.

Discussion

17.	A particle moves along a straight line `OX`. At a time `t` (in seconds) the distance `x` (in metre) of the particle is given by `x = 40 +12 t - t^3`. How long would the particle travel before coming to rest ?A. 24 mB. 40 mC. 56 mD. 16 m
Answer» Correct Answer - C Distance travelled by the particle is `x = 0 + 12 t - t^3` We know that velocity is rate of change of distance i.e., `v = (dx)/(dt)` `:. v = (d)/(dt) (40 + 12t - t^3) = 0 + 12 - 3t^2` but final velocity `v = 0` `12 = 3t^2 = 0` or `t^2 = (12)/(3) = 4` or `t = 2 s` Hence, distance travelled by the particle before coming to rest is given by `x = 40 + 12(2) - (2)^3 = 56 m`.

Discussion

18.	The motion of a particle along a straight line is described by equation : `x = 8 + 12 t - t^3` where `x` is in metre and `t` in second. The retardation of the particle when its velocity becomes zero is.A. `12 ms^-2`B. `24 ms^-2`C. zeroD. `6 ms^-2`
Answer» Correct Answer - A `v = (dx)/(dt) = 12 - 3t^2 = 0`…(i) If velocity is zero, `12 - 3t^2 = 0` which gives `t = 2 sec` For acceleration again differential equation (i) `a = (d^2 x)/(dt^2) = - 6t` …(ii) At time `t = 2 sec, a = -6 xx 2 = -12 m//s^2` Hence retardation `= 12 m//s^2`.

Discussion

19.	A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point. .A. BB. CC. DD. A
Answer» Correct Answer - B Particle has maximum instantaneous velocity at the point at when the slope is maximum. Therefore, `v_(max) = (dx)/(dt)` = maximum slope From figure it is obvious that an point `C`, slope is maximum, hence at this point velocity is maximum.

Discussion

20.	Assertion: The instantaneous velocity does not depend on instantaneous position vector. Reason: The instantaneous velocity and average velocity of a particle are always same.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true.
Answer» Correct Answer - C As `vec v = (d vec s)/(dt)`, hence statement `1` is correct. Instantaneous velocity is same always to average velocity if the particle is moving with constant velocity.

Discussion

21.	A body is moving with uniform velocity of `8 ms^-1`. When the body just crossed another body, the second one starts and moves with uniform acceleration of `4 ms^-2`. The time after which two bodies meet will be :A. 2 sB. 4 sC. 6 sD. 8 s
Answer» Correct Answer - B Let they meet after time `t`. Then distance traveled by both in time `t` should be same. `s = 8t = (1)/(2) 4t^2 rArr t = 4 s`.

Discussion

22.	Two balls are dropped from heights `h` and `2h` respectively from the earth surface. The ratio of time of these balls to reach the earth is.A. `1 : sqrt(2)`B. `sqrt(2) : 1`C. `2 : 1`D. `1 : 4`
Answer» Correct Answer - A `t = sqrt((2 h)/(g)) rArr (t_1)/(t_2) = sqrt((h_1)/(h_2)) = sqrt((1)/(2)) = (1)/(sqrt(2))`.

Discussion

23.	Assertion: A body having non zero acceleration can have a constant velocity. Reason: Acceleration is the rate of change of velocity.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true.
Answer» Correct Answer - D Acceleration is the rate of change of velocity i.e., `a = (dv)/(dt)`.

Discussion

24.	In `Q.28`, the average velocity during the journey is :A. `(-6 hat i+ 9 hat j) km//h`.B. `(-6 hat i+ 8 hat j) km//h`C. `(-9 hat i+ 12 hat j) km//h`D. None of these
Answer» Correct Answer - B Average velocity, `ltvgt = (bar r)/(t) = ((-3 hat i+ 4 hat j)km)/((30//60)hr)` =`(-6 hat i+ 8 hat j) km//h`.

Discussion

25.	A car completes its journey in a straight line in three equal parts with speeds `v_1,v_2 and v_3` respectively. The average speed `v` is given by :A. `(v_1 + v_2 + v_3)/(3)`B. `3 sqrt(v_1 v_2 v_3)`C. `(1)/(v)=(1)/(v_1)+(1)/(v_2)+(1)/(v_3)`D. `(3)/(v)=(1)/(v_1)+(1)/(v_2)+(1)/(v_3)`
Answer» Correct Answer - D Average speed, `v = ("Total distance")/("Total time")` =`(s)/((s//3)/(v_1)+ (s//3)/(v_2)+(s//3)/(v_3)) =(3)/((1)/(v_1)+(1)/(v_2)+(1)/(v_3))` `rArr (3)/(v)=(1)/(v_1)+(1)/(v_2)+(1)/(3)`.

Discussion

26.	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 the 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. `(t_1 t_2)/(t_2 - t_1)`B. `(t_1 t_2)/(t_2 + t_1)`C. `t_1 - t_2`D. `(t_1 + t_2)/(2)`
Answer» Correct Answer - B Velocity of Preeti w.r.t. elevator `= (l)/(t_1)` Velocity of elevator w.r.t. ground `= (l)/(t_2)` Then velocity of girl w.r.t. ground `vec v_("preeti") = vecv_("preeti,elevator") + vecv_("elevator")` `(l)/(t) = (l)/(t_1) + (l)/(t_2)` `rArr t = (l)/(l/(t_1) +(l)/(t_2)) =(t_1 t_2)/(t_1 + t_2)`.

Discussion

27.	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 the moving escalator will be.A. `(t_1 + t_2)//(2)`B. `t_1 t_2//(t_2 - t_1)`C. `t_1 t_2//(t_2 + t_1)`D. `t_1 - t_2`
Answer» Correct Answer - C We have to find net velocity with respect to the Earth that will be equal to velocity o fthe girl plus velocity of escalator. Let dispalcemet is `L`, then Velocity of girl `v_g = (L)/(t_1)` Velocity of escalator `v_e = (L)/(t_2)` Net velocity of the girl `=v_g + v_e = (L)/(t_1)+(L)/(t_2)` It `t` is total time taken is covering distance `L`, then `(L)/(t) = (L)/(t_1)+(L)/(t_2) rArr t = (t_ 1 t_2)/(t_1 + t_2)`.

Discussion

28.	A boy standing at the top of a tower of `20 m`. Height drops a stone. Assuming `g = 10 ms^-2` towards north. The average acceleration of the body is.A. `20 m//s`B. `40 m//s`C. `5 m//s`D. `10 m//s`
Answer» Correct Answer - A We have `v = sqrt(2 gh)` =`sqrt(2 xx 10 xx 20) = sqrt(400) = 20 ms^-1`.

Discussion

29.	Figure shows the position of a particle moving on the x-axis as a function of time, Choose the WRONG statement. A. The particle has come to rest `4` times.B. The maximum speed is at `t = 4 s`C. The average velocity is zero for `t = 2 s` to `t = 6 s`D. Motion of particle is non-uniformly accelerated.
Answer» Correct Answer - B Speed at `t = 4 s` is zero, not maximum.

Discussion

30.	A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. .B. .C. .D. .
Answer» Correct Answer - A For the given condition initial height `h = d` and velocity of the ball is zero. When the ball moves downward its velocity increases and it will be maximum when the ball hits the ground & just after the collision it becomes half and in opposite direction. As the ball moves upward its velocity again decreases and becomes zero at height `d//2`. This explanation match with graph (A).

Discussion

31.	An object is dropped from a height `h`. Then the distance travelled in times `t,2t,3t` are in the ratio.A. `1 : 2 : 3`B. `1 : 4 : 9`C. `1 : 3 : 5`D. `1 : 9 : 5`
Answer» Correct Answer - B `s_1 = (1)/(2) g t^2` `s_2 = (1)/(2) g(2t)^2 = 4 s_1` `s_3 = (1)/(2) g(3t)^2 = 9 s_1` `s_1 : s_2 : s_3 = s_1 : 4 s_1 : 9 s_1 = 1 : 4 : 9`.

Discussion

32.	The velocity-time graph of a body is given below. Find the average velocity from `t = 0` to `t = 40 s`. A. `20 m//s`B. `40 m//s`C. `50 m//s`D. `60 m//s`
Answer» Correct Answer - A Displacement = area upto `t = 40 s` =`(1)/(2) xx 20 xx 20 +(30 - 20) xx 20` `+ (1)/(2)[20 + 60][40 - 30] = 800 m` :. `v_(av) = ("Displacement")/("time") = (800)/(40) = 20 m//s`.

Discussion

33.	The velocity-time graph of a body moving along straight line is as follows : The displacement of the body in 5 s is A. 5 mB. 2 mC. 4 mD. 3 m
Answer» Correct Answer - D Displacement is area under `v-t` graph : =`(1)/(2) xx 3 xx 2 + (1)/(2) xx (1) xx (-2) + 1 xx 1 = 3 m`.

Discussion

34.	A particle starts from rest and undergoes an acceleration as shown in figure. The velocity-time graph from figure will have a shape. A. .B. .C. .D. .
Answer» Correct Answer - A Till `t = 2 s`, there is constant acceleration `a m//s^2` and after that there is constant deceleration `- a m//s` same magnitude. So velocity first increases from `0` to maximum and finally becomes zero.

Discussion

35.	The velocity-time graph of a linear motion os shown figure. The distance from the starting point after `8 sec` is. A. 5 mB. 16 mC. 8 mD. 19 m
Answer» Correct Answer - A Distance from the starting point is equal to displacement. Displacement = area of `v-t` graph =`(1)/(2)[4 + 2]4 - (1)/(2)[4 + 3] xx 2 = 5 m` Here distance travelled is not asked, but the distance from the the starting point.

Discussion

36.	Figure gives the velocity-time graph. This shows that the body is. A. starting from rest and moving with uniform velocityB. moving with uniform retardationC. moving with uniform accelerationD. having same initial and final velocity.
Answer» Correct Answer - B Slope of graph is constant and negative.

Discussion

37.	A lift is going up. The variation in the speed of the lift is as given in the graph in the graph. What is the height to which the lift takes the passengers ? A. `3.6 m`B. `28.8 m`C. `36.0 m`D. Cannot be calculate from the above graph.
Answer» Correct Answer - C Area of trapezium `= (1)/(2) xx 3.6 xx (12 + 8) = 36.0 m`.

Discussion

38.	Assertion: Retardation is directly opposite to the velocity. Reason: Retardation is equal to the time rate of decrease of speed.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - B Retardation is the time rate of decrease of speed hence it is directly opposite to velocity.

Discussion

39.	The velocity of a bullet is reduced from `200 m//s` to `100 m//s` while travelling through a wooden block of thickness `10 cm`. The retardation, assuming it to be uniform, will be.A. `10 xx 10^4 m//s^2`B. `12 xx 10^4 m//s^2`C. `12.5 xx 10^4 m//s^2`D. `15 xx 10^4 m//s^2`
Answer» Correct Answer - D `u = 200 m//s, v = 100 m//s, s = 0.1 m` `a = (u^2 - v^2)/(2 s)` =`((200)^2 -(100)^2)/(2 xx 0.1) =15 xx 10^4 m//s^2`.

Discussion

40.	In the above question what are the distance travelled by each of them till they meet ?A. 30 m and 70 mB. 40 m and 60 mC. 50 m by eachD. 45 m and 55 m
Answer» Correct Answer - B `x = 2 xx 20 = 40 m` `100 - x = 100 - 40 = 60 m`.

Discussion

41.	Assertion: `x-t` graph, for a particle undergoing rectilinear motion, can be as shown in the figure. Reason: Infinitesimal changes in velocity are physically possible only in infinitesimal time. A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true.
Answer» Correct Answer - D At the high point of the graph velocity (slope of `x-t` curve) charges suddenly from a positive to a negative value. This is physically impossible.

Discussion

42.	Assertion: Three projectiles are moving in different paths in the air. Vertical component of relative velocity between any of the pair does not change with time as long as they are in air. Neglect the effect of air friction. Reason: Relative acceleration between any of the pair of projectile is zero.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true.
Answer» Correct Answer - A Acceleration of each of the projectile `= vec g`. Relative acceleration `vec a_r = vec g - vec g = 0`.

Discussion

43.	Refer `Q.107`. Find the displacement of bolt relative to `B` when it hits the flooe of `A`.A. 20 mB. 24 mC. 32 mD. 36 m
Answer» Correct Answer - B Displacement of bolt at `4 s` : `S_b = 20 xx 4 - (1)/(2)10 (4)^2 = 0 m` Displacement of ligt `B` at `4 s` : `S_B = 10 xx 4 - (1)/(2) xx 2 xx (4)^2 = 24 m` downward So displacement of bolt relative to `B` : `S_(b//B) = 0 - 24 = -24 m`.

Discussion

44.	Refer `Q.107`. What is the velocity of bolt relative to lift `B` when it hits the floor of `A` ?A. `18 m//s`B. `20 m//s`C. `22 m//s`D. none of these
Answer» Correct Answer - A Velocity of bolt at `4 s` : `v_b = 20 - g xx 4 = -20 m//s ` =`20 m//s` downward Velocity of lift `B` at `4 s` : `v_B = 10 - 2 xx 4 = 2 m//s` downward. So relative velocity =`v_(b//B) = 20 - 2 = 18 m//s`.

Discussion

45.	A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution between the ball and the floor is `(1)/(2)`. The displacement time graph of the ball will be.A. .B. .C. .D. .
Answer» Correct Answer - C The ball will stop after a long time. The final displacement of the ball will be equal to the height. The motion is first accelerated, then retarded, then accelerated and so on. Hence the correct graph is ( c).

Discussion

46.	The speed -time graph of the ball in the above situation is.A. .B. .C. .D. .
Answer» Correct Answer - B Speed can never be negative. In every collision speed remains half. Motion is accelerated, then retarded, then accelerated and so on. Hence the correct graph is (b).

Discussion

47.	The ratio of the numerical values of the average velocity and average speed of a body is always.A. UnityB. Unity or lessC. Unity or moreD. Less than unity
Answer» Correct Answer - B `(\|"Average velocity"\|)/(\|"Average speed"\|) = (\|"displacement"\|)/(\|"distance"\|) le 1` Dispalcement will either be equal or less than distance. It can never be greater than distance.

Discussion

48.	The speed `v` of a car moving on a straight road changes according to equation, `v^2 = 1 + bx`, where `a and b` are positive constants. Then the magnitude of acceleration in the course of such motion : (x is the distance travelled).A. increasesB. decreasesC. stay constantD. first decreases and then increases
Answer» Correct Answer - C `v^2 = a + bx` `v` increase as `x` increases `a = v (dv)/(dx) = (1)/(2) (d)/(dx) (v^2) = (b)/(2)` Hence, acceleration is constant.

Discussion

49.	When a body is acclerated : (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change. Which of the following is correct ?A. (i) and (ii)B. (i) and (iv)C. (i) onlyD. (ii) and (iii)
Answer» Correct Answer - B Acceleration means change in velocity. So if there is acceleration, velocity will change definitely. Under acceleration, speed mat not change if direction changes and direction may not change if speed changes.

Discussion

50.	Assertion: Area under velocity-time graph give displacement. Reason: Area under acceleration-time graph give velocity.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assersion is false but reason is true.
Answer» Correct Answer - C Area under acceleration-time graph gives change in velocity and not velocity.

Discussion

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Speeds of two identical cars are u and 4u at at specific instant. The ratio of the respective distances in which the two cars are stopped from that instant isA. `1 : 1`B. `1 : 4`C. `1 : 8`D. `1 : 16`

A body `A` starts from rest with an acceleration `a_1`. After `2` seconds, another body `B` starts from rest with an acceleration `a_2`. If they travel equal distances in the `5th` second, after the start of `A`, then the ratio `a_1 : a_2` is equal to :A. `5 : 9`B. `5 : 7`C. `9 : 5`D. `9 : 7`

A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(2v(n + 1))/(n)`B. `(v(n + 1))/(n)`C. `(v(n - 1))/(n)`D. `(2 v(n - 1))/(n)`

A particle starts from rest with uniform acceleration and is velocity after `n` seconds `v`. The displacement of the body in last two seconds is.A. `(v(n + 1))/(n)`B. `(2 v(n + 1))/(n)`C. `(2 v(n - 1))/(n)`D. `(v(n - 1))/(n)`

A body starting from rest moving with uniform acceleration has a displacement of `16 m` in first `4` seconds and `9 m` in first `3` seconds. The acceleration of the body is :A. `1 ms^-2`B. `2 ms^-2`C. `3 ms^-2`D. `4 ms^-2`

A body is moving from rest under constant acceleration and let `S_1` be the displacement in the first `(p - 1)` sec and `S_2` be the displacement in the first `p` sec. The displacement in `(p^2 - p + 1)` sec. will beA. `S_1 + S_2`B. `S_1 S_2`C. `S_1 - S_2`D. `S_1//S_2`

A particle moves along a staight line such that its displacement at any time t is given by `s=t^3-6t^2+3t+4m`. Find the velocity when the acceleration is 0.A. `3 ms^-1`B. `-12 ms^-1`C. `42 ms^-1`D. `-9 ms^-1`

The displacement of a particle moving along x-axis is given by : `x = a + bt + ct^2` The acceleration of the particle is.A. bB. cC. `b + c`D. 2 c

If the velocity of a particle is `v = At + Bt^2`, where `A` and `B` are constant, then the distance travelled by it between `1 s` and `2 s` is :A. `(3)/(2) A + 4B`B. `3 A + 7 B`C. `(3)/(2)A + (7)/(3) B`D. `(A)/(2) + (B)/(3)`

The displacement of a particle is given by `y = a + bt + ct^2 - dt^4`. The initial velocity and acceleration are respectively.A. `b, -4d`B. `-b, 2c`C. `b,2c`D. `2c, -4d`

Refer `Q.139`. The average acceleration of the body from `t = 0` to `t = 15 s` is.A. `1.25 m//s^2`B. `4//7 m//s^2`C. `5//6 m//s^2`D. `7//6 m//s^2`

Refer `Q.149`. At time `t_3`. Which car is moving faster ?A. car `A`B. car `B`C. same speedD. None of these

A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight areA. `(vec v)/(2) and (v)/(2)`B. `0 and (v)/(2)`C. `0 and 0`D. `(vec v)/(2) and 0`.

A car moves from `X` to `Y` with a uniform speed `v_u` and returns to `Y` with a uniform speed `v_d`. The average speed for this round trip is :A. `(2 v_d v_u)/(v_d + v_u)`B. `sqrt(v_u v_d)`C. `(v_d v_u)/(v_d + v_u)`D. `(v_u + v_d)/(2)`

A ball is upward with such a velocity `v` that it returns to the thrower after after `3 s`. Take `g = 10 ms^-2`. Find the value of `v`.A. 15 m//sB. 20 m//sC. 10 m//sD. 5 m//s

A particle moves along a straight line `OX`. At a time `t` (in seconds) the distance `x` (in metre) of the particle is given by `x = 40 +12 t - t^3`. How long would the particle travel before coming to rest ?A. 24 mB. 40 mC. 56 mD. 16 m

The motion of a particle along a straight line is described by equation : `x = 8 + 12 t - t^3` where `x` is in metre and `t` in second. The retardation of the particle when its velocity becomes zero is.A. `12 ms^-2`B. `24 ms^-2`C. zeroD. `6 ms^-2`

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point. .A. BB. CC. DD. A

A body is moving with uniform velocity of `8 ms^-1`. When the body just crossed another body, the second one starts and moves with uniform acceleration of `4 ms^-2`. The time after which two bodies meet will be :A. 2 sB. 4 sC. 6 sD. 8 s

Two balls are dropped from heights `h` and `2h` respectively from the earth surface. The ratio of time of these balls to reach the earth is.A. `1 : sqrt(2)`B. `sqrt(2) : 1`C. `2 : 1`D. `1 : 4`

In `Q.28`, the average velocity during the journey is :A. `(-6 hat i+ 9 hat j) km//h`.B. `(-6 hat i+ 8 hat j) km//h`C. `(-9 hat i+ 12 hat j) km//h`D. None of these

A car completes its journey in a straight line in three equal parts with speeds `v_1,v_2 and v_3` respectively. The average speed `v` is given by :A. `(v_1 + v_2 + v_3)/(3)`B. `3 sqrt(v_1 v_2 v_3)`C. `(1)/(v)=(1)/(v_1)+(1)/(v_2)+(1)/(v_3)`D. `(3)/(v)=(1)/(v_1)+(1)/(v_2)+(1)/(v_3)`

A boy standing at the top of a tower of `20 m`. Height drops a stone. Assuming `g = 10 ms^-2` towards north. The average acceleration of the body is.A. `20 m//s`B. `40 m//s`C. `5 m//s`D. `10 m//s`

A ball is dropped vertically from `a` height `d` above the ground . It hits the ground and bounces up vertically to a height ` (d)//(2). Neglecting subsequent motion and air resistance , its velocity `v` varies with the height `h` above the ground asA. .B. .C. .D. .

An object is dropped from a height `h`. Then the distance travelled in times `t,2t,3t` are in the ratio.A. `1 : 2 : 3`B. `1 : 4 : 9`C. `1 : 3 : 5`D. `1 : 9 : 5`

The velocity-time graph of a body is given below. Find the average velocity from `t = 0` to `t = 40 s`. A. `20 m//s`B. `40 m//s`C. `50 m//s`D. `60 m//s`

The velocity-time graph of a body moving along straight line is as follows : The displacement of the body in 5 s is A. 5 mB. 2 mC. 4 mD. 3 m

A particle starts from rest and undergoes an acceleration as shown in figure. The velocity-time graph from figure will have a shape. A. .B. .C. .D. .

The velocity-time graph of a linear motion os shown figure. The distance from the starting point after `8 sec` is. A. 5 mB. 16 mC. 8 mD. 19 m

Figure gives the velocity-time graph. This shows that the body is. A. starting from rest and moving with uniform velocityB. moving with uniform retardationC. moving with uniform accelerationD. having same initial and final velocity.

A lift is going up. The variation in the speed of the lift is as given in the graph in the graph. What is the height to which the lift takes the passengers ? A. `3.6 m`B. `28.8 m`C. `36.0 m`D. Cannot be calculate from the above graph.

The velocity of a bullet is reduced from `200 m//s` to `100 m//s` while travelling through a wooden block of thickness `10 cm`. The retardation, assuming it to be uniform, will be.A. `10 xx 10^4 m//s^2`B. `12 xx 10^4 m//s^2`C. `12.5 xx 10^4 m//s^2`D. `15 xx 10^4 m//s^2`

In the above question what are the distance travelled by each of them till they meet ?A. 30 m and 70 mB. 40 m and 60 mC. 50 m by eachD. 45 m and 55 m

Refer `Q.107`. Find the displacement of bolt relative to `B` when it hits the flooe of `A`.A. 20 mB. 24 mC. 32 mD. 36 m

Refer `Q.107`. What is the velocity of bolt relative to lift `B` when it hits the floor of `A` ?A. `18 m//s`B. `20 m//s`C. `22 m//s`D. none of these

A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution between the ball and the floor is `(1)/(2)`. The displacement time graph of the ball will be.A. .B. .C. .D. .

The speed -time graph of the ball in the above situation is.A. .B. .C. .D. .

The ratio of the numerical values of the average velocity and average speed of a body is always.A. UnityB. Unity or lessC. Unity or moreD. Less than unity

When a body is acclerated : (i) its velocity always changes (ii) its speed always changes (iii) its direction always changes (iv) its speed may or may not change. Which of the following is correct ?A. (i) and (ii)B. (i) and (iv)C. (i) onlyD. (ii) and (iii)