JEE Physics Motion In A Straight Line MCQs Set 22

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Chapter-wise Objective Questions: Motion In A Straight Line

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Multiple Choice Questions

Question 1: A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speeds of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during the whole motion is:
(a) 4.0 m/s
(b) 5.0 m/s
(c) 5.5 m/s
(d) 4.8 m/s
Show Answer & Explanation

Answer: (a) 4.0 m/s

Explanation:
1. Speed over the second half = average of 4.5 and 7.5 (equal time intervals) = 6 m/s.
2. Let the total distance be 2d. Time = d/3 + d/6 = d/2.
3. Average speed = 2d / (d/2) = 4.0 m/s.

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Question 2: A projectile is fired vertically upwards with initial velocity u. After an interval of T seconds a second projectile is fired vertically upwards, also with initial velocity u. They meet at a time t (measured from the first launch) and at a height h given by:
(a) \( t=\frac{u}{g},\; h=\frac{u^2}{2g}+\frac{gT^2}{8} \)
(b) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}-\frac{gT^2}{8} \)
(c) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}+\frac{gT^2}{8} \)
(d) They never meet
Show Answer & Explanation

Answer: (b) \( t=\frac{u}{g}+\frac{T}{2},\; h=\frac{u^2}{2g}-\frac{gT^2}{8} \)

Explanation:
1. Equal heights: \( ut-\frac{1}{2}gt^2 = u(t-T)-\frac{1}{2}g(t-T)^2 \).
2. Simplifying gives \( uT=\frac{1}{2}g(2tT-T^2) \), so \( t=\frac{u}{g}+\frac{T}{2} \).
3. Substituting back, \( h=ut-\frac{1}{2}gt^2=\frac{u^2}{2g}-\frac{gT^2}{8} \).

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Question 3: A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3 m higher than A, with speed v/2. The maximum height reached by the stone above point B is:
(a) 2 m
(b) 3 m
(c) 1 m
(d) 5 m
Show Answer & Explanation

Answer: (c) 1 m

Explanation:
1. Between A and B: \( v^2-(v/2)^2=2g(3) \), so \( \frac{3v^2}{4}=6g \) and \( v^2=8g \).
2. Above B the stone rises \( h=\frac{(v/2)^2}{2g}=\frac{v^2}{8g} \).
3. Therefore the maximum height above B is 1 m.

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Question 4: A person can swim in still water at 3 km/h. He wants to cross a river flowing at 5 km/h and 120 m wide so that the path travelled is minimum. Which of the following is/are correct?
(A) He must swim making an angle of 127° with the direction of flow of the river.
(B) He must swim perpendicular to the flow of the river.
(C) The length of the shortest possible path is 200 m.
(D) The length of the shortest possible path is 120 m.
(a) B and C only
(b) A and D only
(c) B and D only
(d) A and C only
Show Answer & Explanation

Answer: (d) A and C only

Explanation:
1. Swimmer speed (3) is less than river speed (5), so the shortest path is found when the swimming direction is perpendicular to the resultant velocity.
2. Then \( \cos\theta=-\frac{3}{5} \), so \( \theta=127^\circ \) with the flow.
3. Resultant speed \( =\sqrt{5^2-3^2}=4 \) km/h; cross-stream speed \( =3\sin127^\circ=2.4 \) km/h.
4. Path length \( =120\times\frac{4}{2.4}=200 \) m. So A and C are correct.

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Question 5: From a tower of height H, a particle is thrown vertically upwards with speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:
(a) \( 2gH=nu^2(n-2) \)
(b) \( gH=(n-2)^2u^2 \)
(c) \( 2gH=n^2u^2 \)
(d) \( gH=(n-2)\,2u^2 \)
Show Answer & Explanation

Answer: (a) \( 2gH=nu^2(n-2) \)

Explanation:
1. Time to highest point = u/g, so total time \( t=\frac{nu}{g} \).
2. Displacement is \( -H=ut-\frac{1}{2}gt^2=\frac{nu^2}{g}-\frac{n^2u^2}{2g} \).
3. Hence \( 2gH=n^2u^2-2nu^2=nu^2(n-2) \).

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Question 6: The instantaneous velocity and acceleration of a particle moving in a plane are \( \vec v \) and \( \vec a \). The rate of change of its speed, dv/dt, is equal to which of the following?
(A) \( |\vec a| \)
(B) \( \frac{\vec v\cdot\vec a}{v} \)
(C) the component of \( \vec a \) parallel to \( \vec v \)
(D) the component of \( \vec a \) perpendicular to \( \vec v \)
(a) A and B only
(b) B and C only
(c) A and D only
(d) D only
Show Answer & Explanation

Answer: (b) B and C only

Explanation:
1. Speed squared: \( v^2=\vec v\cdot\vec v \), so \( 2v\frac{dv}{dt}=2\,\vec v\cdot\vec a \).
2. Hence \( \frac{dv}{dt}=\frac{\vec v\cdot\vec a}{v} \), which is option B.
3. This is the projection of \( \vec a \) along \( \vec v \), which is option C. Options A and D are not equal to dv/dt in general.

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Question 7: The initial acceleration of a particle moving in a straight line is a0 and its initial velocity is zero. The acceleration reduces continuously to half in every t0 seconds. The terminal velocity of the particle is:
(a) \( a_0t_0\,\ell n(2) \)
(b) \( a_0t_0 \)
(c) \( \frac{a_0t_0}{\ell n(2)} \)
(d) \( \frac{a_0t_0}{2} \)
Show Answer & Explanation

Answer: (c) \( \frac{a_0t_0}{\ell n(2)} \)

Explanation:
1. The acceleration decays as \( a=a_0\,2^{-t/t_0}=a_0e^{-(\ln 2)t/t_0} \).
2. Terminal velocity \( =\int_0^\infty a\,dt=a_0\cdot\frac{t_0}{\ln 2} \).

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Question 8: Pick the correct statements:
(A) The average speed of a particle in a given time is never less than the magnitude of the average velocity.
(B) It is possible to have a situation in which \( \left|\frac{d\vec v}{dt}\right|\ne 0 \) but \( \frac{d}{dt}|\vec v|=0 \).
(C) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.
(D) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval (infinite accelerations are not allowed).
(a) A and B only
(b) B, C and D only
(c) C and D only
(d) A, B and C only
Show Answer & Explanation

Answer: (d) A, B and C only

Explanation:
1. A: distance is at least the magnitude of displacement, so average speed is at least |average velocity|. True.
2. B: uniform circular motion has non-zero acceleration but constant speed. True.
3. C: one full circle gives zero average velocity while the velocity is never zero. True.
4. D: on a straight line, zero displacement needs the particle to turn back, and at the turning point the velocity is zero. False.

Teacher's Note:
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192. .

Question 9: The acceleration of a particle which is at rest at x = 0 is \( \vec a=(4-2x)\hat i \). Select the correct statements:
(A) The particle comes to rest again at x = 4.
(B) The particle oscillates about x = 2.
(C) The maximum speed of the particle is 4 units.
(a) A and B only
(b) B and C only
(c) A and C only
(d) A, B and C
Show Answer & Explanation

Answer: (a) A and B only

Explanation:
1. \( v\frac{dv}{dx}=4-2x \) gives \( \frac{v^2}{2}=4x-x^2 \) (from rest at x = 0).
2. v = 0 again at x = 4, so A is true. The acceleration is zero at x = 2, which is the centre of the motion between 0 and 4, so B is true.
3. At x = 2: \( v^2=2(8-4)=8 \), so \( v_{max}=2\sqrt2 \), not 4. C is false.

Teacher's Note:
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Question 10: A particle starts from rest and traverses a distance \( \ell \) with uniform acceleration, then moves uniformly over a further distance \( 2\ell \), and finally comes to rest after moving a further distance \( 3\ell \) under uniform retardation. Assuming the entire motion to be rectilinear, the ratio of the average speed over the journey to the maximum speed on its way is:
(a) 1/5
(b) 3/5
(c) 2/5
(d) 4/5
Show Answer & Explanation

Answer: (b) 3/5

Explanation:
1. Let the maximum speed be v. Acceleration stage: average speed v/2, time \( \frac{2\ell}{v} \).
2. Uniform stage: time \( \frac{2\ell}{v} \). Retardation stage: average speed v/2, time \( \frac{6\ell}{v} \).
3. Total time \( \frac{10\ell}{v} \), total distance \( 6\ell \), so average speed \( =0.6v \).

Teacher's Note:
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144. .

Question 11: The acceleration of a particle starting from rest decreases linearly with time from 10 m/s² at t = 0 to zero at t = 11 s, after which it moves with constant velocity. The maximum velocity attained by the particle is:
(a) 500 m/s
(b) 110 m/s
(c) 55 m/s
(d) 650 m/s
Show Answer & Explanation

Answer: (c) 55 m/s

Explanation:
1. The change in velocity equals the area under the a-t graph.
2. Area of the triangle \( =\frac{1}{2}\times 11\times 10=55 \) m/s.

Teacher's Note:
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Question 12: A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s² due west. The distance covered by the particle in the fifth second of its motion is:
(a) 0
(b) 2 m
(c) 4.5 m
(d) 0.5 m
Show Answer & Explanation

Answer: (d) 0.5 m

Explanation:
1. The velocity becomes zero at \( t=4.5 \) s, after which the particle returns.
2. Velocity at t = 4 s is 1 m/s, so it moves \( \frac{1^2}{2\times2}=0.25 \) m further to the east before stopping.
3. From 4.5 s to 5 s it moves back \( \frac{1}{2}(2)(0.5)^2=0.25 \) m.
4. Distance covered \( =0.25+0.25=0.5 \) m (the displacement in this second is zero).

Teacher's Note:
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Question 13: A ball of weight W is thrown upward with a velocity v. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower is:
(a) \( v\sqrt{\frac{W-F}{W+F}} \)
(b) \( v\sqrt{\frac{W}{W+F}} \)
(c) \( v\sqrt{\frac{W}{W-F}} \)
(d) \( v\sqrt{\frac{W+F}{W}} \)
Show Answer & Explanation

Answer: (a) \( v\sqrt{\frac{W-F}{W+F}} \)

Explanation:
1. Going up, the retardation is \( a_1=\frac{(W+F)g}{W} \), so \( h=\frac{v^2}{2a_1} \).
2. Coming down, the acceleration is \( a_2=\frac{(W-F)g}{W} \), so \( v'^2=2a_2h=v^2\frac{W-F}{W+F} \).
3. Thus \( v'=v\sqrt{\frac{W-F}{W+F}} \).

Teacher's Note:
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Question 14: Two balls are dropped to the ground from different heights. One ball is dropped 2 s after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is (g = 10 m/s²):
(a) 20 m
(b) 80 m
(c) 170 m
(d) 40 m
Show Answer & Explanation

Answer: (b) 80 m

Explanation:
1. First ball: \( h_1=\frac{1}{2}(10)(5)^2=125 \) m.
2. Second ball falls for 3 s: \( h_2=\frac{1}{2}(10)(3)^2=45 \) m.
3. Difference \( =125-45=80 \) m.

Teacher's Note:
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134. .

Question 15: A point moves in a straight line so that its displacement x at time t is given by \( x^2=1+t^2 \). Its acceleration at any time t is (select all correct expressions):
(A) \( \frac{1}{x^3} \)
(B) \( \frac{-t}{x^3} \)
(C) \( \frac{1}{x}-\frac{t^2}{x^3} \)
(D) \( \frac{1}{x}-\frac{1}{x^2} \)
(a) A and B only
(b) B and D only
(c) A and C only
(d) C and D only
Show Answer & Explanation

Answer: (c) A and C only

Explanation:
1. Differentiate: \( 2x\dot x=2t \), so \( v=\frac{t}{x} \).
2. \( a=\frac{dv}{dt}=\frac{1}{x}-\frac{t}{x^2}\dot x=\frac{1}{x}-\frac{t^2}{x^3} \), which is option C.
3. Also \( \frac{1}{x}-\frac{t^2}{x^3}=\frac{x^2-t^2}{x^3}=\frac{1}{x^3} \), which is option A.

Teacher's Note:
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Question 16: 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) \( -2\beta^2x^{-2n+1} \)
(b) \( -2n\beta^2x^{-2n-1} \)
(c) \( -2n\beta^2e^{-4n+1} \)
(d) \( -2n\beta^2x^{-4n-1} \)
Show Answer & Explanation

Answer: (d) \( -2n\beta^2x^{-4n-1} \)

Explanation:
1. \( a=v\frac{dv}{dx} \).
2. \( \frac{dv}{dx}=-2n\beta x^{-2n-1} \).
3. \( a=\beta x^{-2n}\times(-2n\beta x^{-2n-1})=-2n\beta^2x^{-4n-1} \).

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Question 17: A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2 and h3 is:
(a) \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \)
(b) \( h_2=3h_1\text{ and }h_3=3h_2 \)
(c) h1 = h2 = h3
(d) \( h_1=2h_2=3h_3 \)
Show Answer & Explanation

Answer: (a) \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \)

Explanation:
1. Starting from rest, distances in successive equal intervals are in the ratio 1 : 3 : 5.
2. So \( h_2=3h_1 \) and \( h_3=5h_1 \), giving \( h_1=\frac{h_2}{3}=\frac{h_3}{5} \).

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Question 18: The motion of a particle along a straight line is described by the equation \( x=8+12t-t^3 \), where x is in metres and t in seconds. The retardation of the particle when its velocity becomes zero is:
(a) 24 m/s²
(b) 12 m/s²
(c) zero
(d) 6 m/s²
Show Answer & Explanation

Answer: (b) 12 m/s²

Explanation:
1. \( v=\frac{dx}{dt}=12-3t^2 \), which is zero at t = 2 s.
2. \( a=\frac{dv}{dt}=-6t \), so at t = 2 s, \( a=-12 \) m/s².
3. Retardation = 12 m/s².

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Question 19: A particle moves along a straight line according to \( x=(t+5)^{-1} \). The acceleration of the particle is proportional to:
(a) \( (\text{distance})^{2} \)
(b) \( (\text{distance})^{-2} \)
(c) \( (\text{velocity})^{3/2} \)
(d) \( (\text{velocity})^{2/3} \)
Show Answer & Explanation

Answer: (c) \( (\text{velocity})^{3/2} \)

Explanation:
1. \( v=-(t+5)^{-2}=-x^2 \), so \( |v|=x^2 \).
2. \( a=2(t+5)^{-3}=2x^3 \).
3. Since \( x=|v|^{1/2} \), \( a=2|v|^{3/2} \), so a is proportional to (velocity)^{3/2}.

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Question 20: A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (g = 10 m/s²)
(a) 55 m/s
(b) 40 m/s
(c) 60 m/s
(d) 75 m/s
Show Answer & Explanation

Answer: (d) 75 m/s

Explanation:
1. First ball falls for 18 s: \( \frac{1}{2}(10)(18)^2=1620 \) m.
2. Second ball moves for 12 s: \( 12v+\frac{1}{2}(10)(12)^2=12v+720 \).
3. Equate: \( 12v+720=1620 \), so v = 75 m/s.

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Motion In A Straight Line Objective Questions & Solutions for JEE Physics

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