JEE Physics Motion In A Straight Line MCQs Set 04

Physics Objective Questions and Answers: Motion In A Straight Line

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Multiple Choice Questions

Question 1: A particle is thrown up vertically with a speed \( v_1 \) in air. It takes time \( t_1 \) in the upward journey and returns to the starting point in time \( t_2\ (>t_1) \) in the downward journey with a speed \( v_2 \). Then
(a) \( v_1=v_2 \)
(b) \( v_1\lt v_2 \)
(c) \( v_1>v_2 \)
(d) data insufficient
Show Answer & Explanation

Answer: (c) \( v_1>v_2 \)

Explanation:
1. Air resistance removes mechanical energy during the whole flight.
2. So the particle returns with less kinetic energy, i.e. a smaller speed: \( v_2\lt v_1 \).

Teacher's Note:
1. Work done by air resistance is always negative.

Question 2: A body is projected vertically upward from the surface of the earth. Taking upward as positive, its velocity-time graph (till it returns to the ground) is
(a) a straight line of negative slope that starts at a positive velocity and crosses the time axis
(b) a straight line of positive slope
(c) a parabola opening downwards
(d) a V shaped graph whose vertex touches the time axis
Show Answer & Explanation

Answer: (a) a straight line of negative slope that starts at a positive velocity and crosses the time axis

Explanation:
1. \( v=u-gt \): a straight line with constant slope \( -g \).
2. It starts at \( +u \), crosses zero at the top and becomes negative on the way down.

Teacher's Note:
1. Constant acceleration gives a straight line in the v-t graph.

Question 3: A body starts from rest and with a uniform acceleration of 10 m/s\( ^2 \) for 5 seconds. During the next 10 seconds it moves with uniform velocity. The total distance travelled is
(a) 100 m
(b) 125 m
(c) 500 m
(d) 625 m
Show Answer & Explanation

Answer: (d) 625 m

Explanation:
1. First 5 s: \( s_1=\frac12(10)(25)=125 \) m and \( v=50 \) m/s.
2. Next 10 s: \( s_2=50\times10=500 \) m.
3. Total \( =625 \) m.

Teacher's Note:
1. Split the motion into two stages.

Question 4: A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at t = 2 s and (13 m, 14 m) at t = 5 s. The average velocity vector from t = 0 to t = 5 s is
(a) \( \frac15(13\hat i+14\hat j) \)
(b) \( \frac73(\hat i+\hat j) \)
(c) \( 2(\hat i+\hat j) \)
(d) \( \frac{11}5(\hat i+\hat j) \)
Show Answer & Explanation

Answer: (d) \( \frac{11}5(\hat i+\hat j) \)

Explanation:
1. Displacement \( =(13-2)\hat i+(14-3)\hat j=11\hat i+11\hat j \).
2. \( \vec v_{av}=\frac{11}5(\hat i+\hat j) \).

Teacher's Note:
1. Average velocity uses only the end points.

Question 5: The position x of a particle with respect to time t along the x-axis is given by \( x=9t^2-t^3 \) (metres, seconds). What will be the position of this particle when it achieves maximum speed along the +x direction?
(a) 24 m
(b) 32 m
(c) 54 m
(d) 81 m
Show Answer & Explanation

Answer: (c) 54 m

Explanation:
1. \( v=18t-3t^2 \) is maximum when \( a=18-6t=0\Rightarrow t=3 \) s.
2. \( x(3)=81-27=54 \) m.

Teacher's Note:
1. Maximum speed occurs where the acceleration is zero.

Question 6: A stone is dropped from a certain height which can reach the ground in 5 seconds. It is stopped after 3 seconds of its fall and is again released. The total time taken by the stone to reach the ground will be
(a) 6 s
(b) 6.5 s
(c) 7 s
(d) 7.5 s
Show Answer & Explanation

Answer: (c) 7 s

Explanation:
1. Total height \( H=\frac12g(25) \). After 3 s it has dropped \( \frac12g(9) \).
2. The remaining height is \( \frac12g(16) \), which takes 4 s to fall from rest.
3. Total time \( =3+4=7 \) s.

Teacher's Note:
1. The remaining distance is found by subtraction.

Question 7: In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity u, the horizontal range is
(a) \( \frac{u^2}g \)
(b) \( \frac{u^2}{2g} \)
(c) \( \frac{2u^2}g \)
(d) \( \frac{u^2}{g^2} \)
Show Answer & Explanation

Answer: (a) \( \frac{u^2}g \)

Explanation:
1. Equal inclination to the horizontal and the vertical means \( \theta=45^\circ \).
2. \( R=\frac{u^2\sin2\theta}g=\frac{u^2}g \).

Teacher's Note:
1. Use the range formula at 45 degrees.

Question 8: A body dropped from a tower reaches the ground in 4 s. The height of the tower is about (g = 10 m/s\( ^2 \))
(a) 80 m
(b) 20 m
(c) 160 m
(d) 40 m
Show Answer & Explanation

Answer: (a) 80 m

Explanation:
1. \( h=\frac12gt^2=\frac12\times10\times16=80 \) m.

Teacher's Note:
1. Use the free fall formula.

Question 9: The velocity-time graph of a body consists of straight segments: velocity falls from 40 m/s at t = 0 to 20 m/s at t = 2 s, stays at 20 m/s until t = 3 s, falls to 0 at t = 4 s, and then falls to -40 m/s at t = 5 s. The distance travelled and the displacement during the 5 seconds (in metres) are
(a) 75, 75
(b) 110, 70
(c) 110, 110
(d) 110, 40
Show Answer & Explanation

Answer: (b) 110, 70

Explanation:
1. Areas: 0 to 2 s: \( \frac{40+20}2\times2=60 \); 2 to 3 s: 20; 3 to 4 s: \( \frac12\times1\times20=10 \); 4 to 5 s: \( \frac12\times1\times40=20 \) (below the axis).
2. Distance \( =60+20+10+20=110 \) m. Displacement \( =60+20+10-20=70 \) m.

Teacher's Note:
1. Add the areas for distance; subtract the area below the axis for displacement.

Question 10: The position of a particle moving along the x-axis is given by \( x=2+8t-4t^2 \). The distance travelled by the particle from t = 0 to t = 2 s is
(a) 0
(b) 8
(c) 12
(d) 16
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. \( v=8-8t=0 \) at \( t=1 \) s, where \( x=2+8-4=6 \).
2. \( x(0)=2,\ x(2)=2 \). Distance \( =(6-2)+(6-2)=8 \) m.

Teacher's Note:
1. Find the turning point before adding distances.

Question 11: Two cars P and Q start from a point at the same time in a straight line and their positions are \( x_P(t)=at+bt^2 \) and \( x_Q(t)=ft-t^2 \). At what time do the cars have the same velocity?
(a) \( \frac{a+f}{2(1+b)} \)
(b) \( \frac{f-a}{2(1+b)} \)
(c) \( \frac{a-f}{1+b} \)
(d) \( \frac{a+f}{2(b-1)} \)
Show Answer & Explanation

Answer: (b) \( \frac{f-a}{2(1+b)} \)

Explanation:
1. \( v_P=a+2bt \) and \( v_Q=f-2t \).
2. Equal velocities: \( a+2bt=f-2t\Rightarrow t=\frac{f-a}{2(1+b)} \).

Teacher's Note:
1. Differentiate and equate the velocities.

Question 12: 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) \( h_1=h_2=h_3 \)
(b) \( h_1=2h_2=3h_3 \)
(c) \( h_1=\frac{h_2}3=\frac{h_3}5 \)
(d) \( h_2=3h_1 \) and \( h_3=3h_2 \)
Show Answer & Explanation

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

Explanation:
1. From rest, equal successive time intervals give distances in the ratio 1 : 3 : 5.
2. So \( h_1=\frac{h_2}3=\frac{h_3}5 \).

Teacher's Note:
1. Use Galileo's odd number rule.

Question 13: A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 m/s to 20 m/s while passing through a distance of 135 m in t seconds. The value of t is
(a) 12
(b) 9
(c) 10
(d) 1.8
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. Average velocity \( =\frac{10+20}2=15 \) m/s.
2. \( t=\frac{135}{15}=9 \) s.

Teacher's Note:
1. Use the average of the initial and final velocities.

Question 14: A rocket is launched upward from the earth's surface. Its velocity-time graph is: velocity rises uniformly from 0 to 1000 m/s in the first 20 s (OA), then falls uniformly through zero at t = 120 s (B) to a negative value at t = 140 s (C). The height covered by the rocket before retardation (while the engine burns) is
(a) 1 km
(b) 10 km
(c) 20 km
(d) 60 km
Show Answer & Explanation

Answer: (b) 10 km

Explanation:
1. Up to t = 20 s: \( \frac12\times20\times1000=10000 \) m = 10 km.

Teacher's Note:
1. Take the area up to the point A.

Question 15: Initially a body is at rest. If its acceleration is 5 m/s\( ^2 \), the distance travelled in the 18th second is
(a) 86.6 m
(b) 87.5 m
(c) 88 m
(d) 89 m
Show Answer & Explanation

Answer: (b) 87.5 m

Explanation:
1. \( s_n=u+\frac a2(2n-1)=\frac52\times35=87.5 \) m.

Teacher's Note:
1. Use the formula for the distance in the nth second.

Question 16: 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) \( 2c,\ -4d \)
Show Answer & Explanation

Answer: (c) \( b,\ 2c \)

Explanation:
1. \( v=b+2ct-4dt^3 \), so \( v(0)=b \).
2. \( a=2c-12dt^2 \), so \( a(0)=2c \).

Teacher's Note:
1. Evaluate the derivatives at t = 0.

Question 17: The velocity-time graph of a moving object is a straight line at 1 m/s from 0 to 20 s, rising to 4 m/s at 30 s, falling back to 1 m/s at 40 s, and then constant at 1 m/s up to 50 s. The total displacement during the interval when there is non-zero acceleration (20 s to 40 s) is
(a) 60 m
(b) 50 m
(c) 30 m
(d) 40 m
Show Answer & Explanation

Answer: (b) 50 m

Explanation:
1. 20 to 30 s: \( \frac{1+4}2\times10=25 \) m.
2. 30 to 40 s: \( \frac{4+1}2\times10=25 \) m.
3. Total \( =50 \) m.

Teacher's Note:
1. The acceleration is non-zero only while the velocity changes.

Question 18: A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in the air for
(a) 12 s
(b) 13 s
(c) 25 s
(d) 26 s
Show Answer & Explanation

Answer: (b) 13 s

Explanation:
1. Distance in the first 5 s \( =\frac12g(25) \).
2. Distance in the nth second \( =\frac g2(2n-1) \).
3. \( 2n-1=25\Rightarrow n=13 \) s.

Teacher's Note:
1. Use the nth second formula.

Question 19: A boy standing at the top of a tower of 20 m height drops a stone. Assuming g = 10 m/s\( ^2 \), the velocity with which it hits the ground is
(a) 10.0 m/s
(b) 20.0 m/s
(c) 40.0 m/s
(d) 5.0 m/s
Show Answer & Explanation

Answer: (b) 20.0 m/s

Explanation:
1. \( v=\sqrt{2gh}=\sqrt{2\times10\times20}=20 \) m/s.

Teacher's Note:
1. Use v squared = 2 g h.

Question 20: A rocket is launched upward from the earth's surface. Its velocity-time graph is: velocity rises uniformly from 0 to 1000 m/s in the first 20 s (OA), then falls uniformly through zero at t = 120 s (B) to a negative value at t = 140 s (C). The acceleration of the rocket during the burn is
(a) 50 m/s\( ^2 \)
(b) 100 m/s\( ^2 \)
(c) 10 m/s\( ^2 \)
(d) 1000 m/s\( ^2 \)
Show Answer & Explanation

Answer: (a) 50 m/s\( ^2 \)

Explanation:
1. Slope of OA \( =\frac{1000}{20}=50 \) m/s\( ^2 \).

Teacher's Note:
1. Use the slope of OA.

Assertion–Reason Questions

Question 21:
Assertion (A): An object can have constant speed but variable velocity.
Reason (R): Speed is a scalar but velocity is a vector quantity.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Velocity changes when its direction changes even if the speed is constant (uniform circular motion).
2. This is possible because velocity is a vector while speed is a scalar.

Teacher's Note:
1. Direction is part of the velocity.

Question 22:
Assertion (A): A positive acceleration can be associated with 'slowing down' of the body.
Reason (R): The origin and the positive direction of an axis are a matter of choice.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. If the body moves in the negative direction while the acceleration is positive, its speed decreases.
2. Which direction is positive is a free choice, which explains why this can happen.

Teacher's Note:
1. Compare the signs of velocity and acceleration.

Question 23:
Assertion (A): Distance is the actual length of the path but the magnitude of displacement is the shortest distance between the initial and final positions.
Reason (R): Distance is a scalar quantity and it is always positive or zero but displacement is a vector quantity. It may be positive, negative or zero.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true.
2. The reason talks about the scalar and vector nature of the two quantities, not about path length versus shortest distance, so it does not explain the assertion.

Teacher's Note:
1. Compare the explanation with the assertion.

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