Question 1
A ball falls from a building and covers 5m in 10s. What is the acceleration?
A. 0.1 m/s2
B. 0.2 m/s2
C. 9.81 m/s2
D. 10 m/s2
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Question 2
A car is travelling in the north direction. To stop, it produces a deceleration of 60 m/s2. Which of the following is a correct representation for the deceleration?
A. 60 m/s2 Northwards
B. 60 m/s2 Southwards
C. 60 m/s2 Eastwards
D. 60 m/s2 Westwards
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Question 3
How does the displacement v/s time graph of a uniformly accelerated motion look like?
A. A straight line
B. A parabola
C. A hyperbola
D. An ellipse
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Question 4
If the velocity varies parabolically, how does the acceleration vary?
A. Linearly
B. Hyperbolically
C. Parabolically
D. Elliptically
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Question 5
The expression for displacement is x = sin(5t). The expression for acceleration is ______
A. 5sin(5t)
B. 25cos(5t)
C. -25sin(5t)
D. -5cost(5t)
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Question 6
The gradient of velocity v/s time graph is equal to____
A. Velocity
B. Acceleration
C. Distance
D. Momentum
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Question 7
The velocity of a body varies as, v = 2t+5t2. What is the acceleration at t = 10?
A. 102
B. 100
C. -100
D. 50
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Question 8
The velocity of a ship varies with time as v = 5t3. What is the acceleration at t = 2?
A. 60
B. 56
C. 40
D. 100
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Question 9
The velocity of a truck changes form 3 m/s to 5 m/s in 5s. What is the acceleration in m/s2?
A. 0.4
B. 0.5
C. 4
D. 5
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Question 10
A uniformly accelerated body has ____
A. Constant speed
B. Constant velocity
C. Constant force
D. Constant momentum
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Question 11
What is negative acceleration known as?
A. Deceleration
B. Acceleration
C. Escalation
D. Relaxation
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Question 12
Which of the following is the correct formula for finding acceleration?
A. a = dx/dt
B. a = d2x/dt2
C. a = x/t
D. a = d3x/dt3
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Question 13
A particle is moving along the straight line OX and its distance x is in metres from O after t seconds from start is given by x = t3 – t2 – 5t. What will be the acceleration of the particle when it is at a distance 28 metres from O?
A. 20 m/sec2
B. 22 m/sec2
C. 24 m/sec2
D. 26 m/sec2
View Answer
Answer: Option B
Explanation:
We have, x = t3 – t2 – 5t ……….(1)When x = 28, then from (1) we get,t3 – t2 – 5t = 28Or t3 – t2 – 5t – 28 = 0Or (t – 4)(t2 + 3t +7) = 0Thus, t = 4Let v and f be the velocity and acceleration respectively of the particle at time t seconds. Then,v = dx/dt = d(t3 – t2 – 5t)/dt= 3t2 – 2t – 5And f = dv/dt = d(3t2 – 2t – 5)/dt= 6t – 2Therefore, the acceleration of the particle at the end of 4 seconds i.e., when the particle is at a distance of 28 metres from O,[f]t = 4 = (6*4 – 2) m/sec2= 22 m/sec2
Question 14
A particle is moving in a straight line and its distance s cm from a fixed point in the line after t seconds is given by s = 12t – 15t2 + 4t3. What is the acceleration of the particle after 3 seconds?
A. 41 cm/sec2
B. 42 cm/sec2
C. 43 cm/sec2
D. 44 cm/sec2
View Answer
Answer: Option B
Explanation:
We have, s = 12t – 15t2 + 4t3 ……….(1)Differentiating both side of (1) with respect to t we get,(ds/dt) = 12 – 30t + 12t2And d2s/dt2 = -30 + 24tSo, acceleration of the particle after 3 seconds is,[d2s/dt2]t = 3 = – 30 + 24(3)= 42 cm/sec2.
Question 15
A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t3 – 12t + 11. What is the acceleration of the particle at the end of 2 seconds?
A. 22 cm/sec2
B. 24 cm/sec2
C. 26 cm/sec2
D. 28 cm/sec2
View Answer
Answer: Option A
Explanation:
We have, x = 2t3 – 12t + 11 ……….(1)Let v and f be the velocity and acceleration respectively of the particle at time t seconds.Then, v = dx/dt = d(2t3 – 12t + 11)/dt= 6t2 – 12 ……….(2)And f = dv/dt = d(6t2 – 12)/dt= 12t ……….(3)Putting the value of t = 2 in (3),Therefore, the displacement of the particle at the end of 2 seconds,12t = 12(2)= 24 cm/sec2
Question 16
A particle is moving in a straight line is at a distance x from a fixed point in the straight line at time t seconds, where x = 2t3 – 12t + 11. What is the average acceleration of the particle at the end of 3 seconds?
A. 28 cm/sec2
B. 30 cm/sec2
C. 32 cm/sec2
D. 26 cm/sec2
View Answer
Answer: Option B
Explanation:
We have, x = 2t3 – 12t + 11 ……….(1)Let v and f be the velocity and acceleration respectively of the particle at time t seconds.Then, v = dx/dt = d(2t3 – 12t + 11)/dt= 6t2 – 12 ……….(2)And f = dv/dt = d(6t2 – 12)/dt= 12t ……….(3)Putting the value of t = 2 in (3),Therefore, the acceleration of the particle at the end of 2 seconds,12t = 12(2)= 24 cm/sec2Now putting the value of t = 2 in (2),We get the displacement of the particle at the end of 2 seconds,6t2 – 12 = 6(2)2 – 12= 12 cm/sec ……….(4)And putting the value of t = 3 in (2),We get the displacement of the particle at the end of 3 seconds,6t2 – 12 = 6(3)2 – 12= 42 cm/sec ……….(5)Thus, change in velocity is, (5) – (4),=42 – 12= 30cm/sec.Thus, the average acceleration of the particle at the end of 3 seconds is,= (change of velocity)/time= (30 cm/sec)/1 sec= 30 cm/sec2
Question 17
A particle moves along the straight-line OX, starting from O with a velocity 4 cm/sec. At time t seconds its acceleration is (5 + 6t) cm/sec2. What will be the distance from O after 4 seconds?
A. 110 cm
B. 120 cm
C. 130 cm
D. 140 cm
View Answer
Answer: Option B
Explanation:
Let, vcm/sec be the velocity and x cm be the distance of the particle from O and time t seconds.Then the velocity of the particle and acceleration at time t seconds is, v = dx/dt and dv/dt respectively.By the question, dv/dt = 5 + 6tOr dv = (5 + 6t) dtOr ∫dv = ∫(5 + 6t) dtOr v = 5t + 6*(t2)/2 + A ……….(1)By question v = 4, when t = 0;Hence, from (1) we get, A = 4.Thus, v = dx/dt = 5t + 3(t2) + 4 ……….(2)Or ∫dx = ∫(5t + 3(t2) + 4) dtOr x = 5t2/2 + t3 + 4t + B ……….(3)By question x = 0, when t = 0;Hence, from (3) we get, B = 0Thus, x = 5t2/2 + t3 + 4tThus, distance of the particle after 4 seconds,= [x]t = 4 = (5/2*42 + 43 + 4*4) [putting t = 4 in (4)] = 120 cm.
Question 18
A particle moves along the straight-line OX, starting from O with a velocity 4 cm/sec. At time t seconds its acceleration is (5 + 6t) cm/sec2. What will be the velocity of the particle from O after 4 seconds?
A. 70 cm/sec
B. 71 cm/sec
C. 72 cm/sec
D. 73 cm/sec
View Answer
Answer: Option C
Explanation:
Let, vcm/sec be the velocity and x cm be the distance of the particle from O and time t seconds.Then the velocity of the particle at time t seconds is, v = dx/dtBy the question, dv/dt = 5 + 6tOr dv = (5 + 6t) dtOr ∫dv = ∫(5 + 6t) dtOr v = 5t + 6*(t2)/2 + A ……….(1)By question v = 4, when t = 0;Hence, from (1) we get, A = 4.Thus, v = dx/dt = 5t + 3(t2) + 4 ……….(2)Thus, velocity of the particle after 4 seconds,= [v]t = 4 = (5*4 + 3*42 + 4) [putting t = 4 in (2)] = 72 cm/sec.
Question 19
A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?
A. 1/v = 1/u + kx
B. 1/v = 1/u – 2kx
C. 1/v = 1/u – kx
D. 1/v = 1/u + 2kx
View Answer
Answer: Option A
Explanation:
Since the particle is moving in a straight line under a retardation kv3, hence, we have,dv/dt = -kv3 ……….(1)Or dv/dx*dx/dt = -kv3Or v(dv/dx) = -kv3 [as, dx/dt = v] Or ∫v-2 dv = -k∫dxOr v-2+1/(-2 + 1) = -kx – B, where B is a integration constantOr 1/v = kx + B ……….(3)Given, v = u, when x = 0; hence, from (3) we get, B = 1/uThus, putting B = 1/u in (3) we get,1/v = 1/u + kx.
Question 20
A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the acceleration of the particle during first 3 seconds after the start?
A. 10cm/sec2
B. 12cm/sec2
C. 14cm/sec2
D. 16cm/sec2
View Answer
Answer: Option C
Explanation:
Let f be the acceleration of the particle in time t seconds. Then,f = dv/dt = d(3t2 – 4t + 5)/dt= 6t – 4 ……….(1)Therefore, the acceleration of the particle at the end of 3 seconds,= [f]t = 3 = (6*3 – 4) cm/sec2= 14cm/sec2
Question 21
A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What is the meaning of the constant c?
A. Uniform acceleration
B. Non – uniform acceleration
C. Uniform retardation
D. Non – uniform retardation
View Answer
Answer: Option A
Explanation:
We have, x = a + bt + ct2 ……….(1)Let, v and f be the velocity and acceleration of a particle at time t seconds.Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct ……….(2)And f = dv/dt = d(b + ct)/dt = c ……….(3)Since f = dv/dt = c, hence, c represents the uniform acceleration of the particle.
Question 22
A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c < 0?
A. Uniform retardation
B. Uniform speed
C. Uniform acceleration
D. Uniform velocity
View Answer
Answer: Option A
Explanation:
We have, x = a + bt + ct2 ……….(1)Let, v and f be the velocity and acceleration of a particle at time t seconds.Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct ……….(2)And f = dv/dt = d(b + ct)/dt = c ……….(3)Clearly, when c < 0,implies f < 0.Hence in this case the particle moves withan uniform retardation.
Question 23
A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c > 0?
A. Uniform retardation
B. Uniform speed
C. Uniform positive acceleration
D. Uniform velocity
View Answer
Answer: Option C
Explanation:
We have, x = a + bt + ct2 ……….(1)Let, v and f be the velocity and acceleration of a particle at time t seconds.Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct ……….(2)And f = dv/dt = d(b + ct)/dt = c ……….(3)Clearly, when c > 0,implies f > 0.Hence in this case the particle moves withan uniform positive acceleration.
Question 24
A particle starts moving from rest with an acceleration in a fixed direction. If its acceleration at time t be(a – bt2),where a and b are positive constants then which one is correct?
A. [v]max = 4a√a/3√b
B. [v]max = 2a√a/3√b
C. [v]max = 2a√a/3√b
D. [v]max = 4a√a/3√b
View Answer
Answer: Option B
Explanation:
If v be the velocity of the moving particle at time t then its acceleration at time t will be dv/dt. By question,dv/dt = a – bt2Integrating we get, v = ∫ a – bt2 dt = at – bt3/3 + k ……….(1)where k is constant of integration.Given, v = 0, when t = 0; hence from (1) we get,0 = a(0) – b/3(0) + kOr k = 0Thus, v = at – bt3/3 ……….(2)Again, d2v/dt2 = d(a – bt2)/dt = -2btNow, for minimum or maximum value of v we have,dv/dt = 0Or a – bt2 = 0Or t2 = a/bOr t = √a/√b [Since t > 0 and a, b are positive constants] At t = √a/√b we have d2v/dt2 = -2b(√a/√b) < 0 [Since t < 0 and a, b are positive constants] Putting t = √a/√b in (2), We find, v is maximum at t = √a/√b and the minimum value of v is, [v]max = 2a√a/3√b.
Question 25
The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle at the instant, then which one is correct?
A. Moves with retardation 2av2
B. Moves with retardation 2av3
C. Moves with acceleration 2av3
D. Moves with acceleration 2av2
View Answer
Answer: Option B
Explanation:
We have, t = ax2 + bx + c ……….(1)Differentiating both sides of (1) with respect to x we get,dt/dx = d(ax2 + bx + c)/dx = 2ax + bThus, v = velocity of the particle at time t= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)-1 ……….(2)Thus, acceleration of the particle at time t is,= dv/dt = d((2ax + b)-1)/dt= -1/(2ax + b)2 * 2av= -v2*2av [as, v = 1/(2ax + b)] = -2av3That is the particle is moving with retardation 2av3.
Question 26
Two straight railway lines meet at right angles. A train starts from the junction along one line and at the same time instant, another train starts towards the junction from a station on the other line and they move at the same uniform velocity.When will they be nearest to each other?
A. When they are equal distance from the junction
B. When they are in unequal distance from the junction
C. When they form a right angle at the junction
D. Data not sufficient
View Answer
Answer: Option A
Explanation:
Let OX and OY be two straight railway lines and they meet at O at right angles.One train starts from the junction and moves with uniform velocity u km/hr along the line OY.And at the same instant, another train starts towards the junction O from station A on the line OX with same uniform velocity u km/hr.Let C and B be the position of the two trains on lines OY and OX respectively after t hours from the start.Then OC = AB = ut km. Join BC and let OA = a km and BC = x km.Then OB = a – ut.Now, from the right angled triangle BOC we get,BC2 = OB2 + OC2Or x2 = (a – ut)2 + (ut)2Thus, d(x2)/dt = 2(a – ut)(-u) + u2(2t)And d2(x2)/dt2 = 2u2 + 2u2 = 4u2For maximum or minimum value of x2(i.e., x) we must have,d(x2)/dt = 0Or 2(a – ut)(-u) + u2(2t) = 0Or 2ut = a [Since u ≠ 0] Or t = a/2uAgain at t = a/2u we have, d2(x2)/dt2 = 4u2 > 0Therefore, x2(i.e., x) is minimum at t = a/2uNow when t= a/2u, then OC = ut = u(a/2u) = a/2 and OB =a – ut = a – u(a/2u) = a/2 that is at t = a/2u we have, OC = OB.Therefore, the trains are nearest to each other when they are equally distant from the junction.