# Class 12 Maths MCQ – Calculus Application – Acceleration

This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Acceleration”.

1. 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

Explanation: We have, s = 12t – 15t2 + 4t3 ……….(1)
Differentiating both side of (1) with respect to t we get,
(ds/dt) = 12 – 30t + 12t2
And d2s/dt2 = -30 + 24t
So, acceleration of the particle after 3 seconds is,
[d2s/dt2]t = 3 = – 30 + 24(3)
= 42 cm/sec2.

2. 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

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

3. 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

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/sec2
Now 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

4. 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

Explanation: We have, x = t3 – t2 – 5t  ……….(1)
When x = 28, then from (1) we get,
t3 – t2 – 5t = 28
Or t3 – t2 – 5t – 28 = 0
Or (t – 4)(t2 + 3t +7) = 0
Thus, t = 4
Let 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 – 5
And f = dv/dt = d(3t2 – 2t – 5)/dt
= 6t – 2
Therefore, 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

5. 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

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.
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6. 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

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.

7. 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

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.

8. 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

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

9. 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

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/dt
By the question, dv/dt = 5 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ∫(5 + 6t) dt
Or 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.

10. 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

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 + 6t
Or dv = (5 + 6t) dt
Or ∫dv = ∫(5 + 6t) dt
Or 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) dt
Or x = 5t2/2 + t3 + 4t + B   ……….(3)
By question x = 0, when t = 0;
Hence, from (3) we get, B = 0
Thus, x = 5t2/2 + t3 + 4t
Thus, distance of the particle after 4 seconds,
= [x]t = 4 = (5/2*42 + 43 + 4*4)  [putting t = 4 in (4)]
= 120 cm.

11. 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

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 = -kv3
Or v(dv/dx) = -kv3  [as, dx/dt = v]
Or ∫v-2 dv = -k∫dx
Or v-2+1/(-2 + 1) = -kx – B, where B is a integration constant
Or 1/v = kx + B   ……….(3)
Given, v = u, when x = 0; hence, from (3) we get, B = 1/u
Thus, putting B = 1/u in (3) we get,
1/v = 1/u + kx.

12. 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

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 + b
Thus, 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)]
= -2av3
That is the particle is moving with retardation 2av3.

13. 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

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 + OC2
Or x2 = (a – ut)2 + (ut)2
Thus, d(x2)/dt = 2(a – ut)(-u) + u2(2t)
And d2(x2)/dt2 = 2u2 + 2u2 = 4u2
For maximum or minimum value of x2(i.e., x) we must have,
d(x2)/dt = 0
Or 2(a – ut)(-u) + u2(2t) = 0
Or 2ut = a   [Since u ≠ 0]
Or t = a/2u
Again at t = a/2u we have, d2(x2)/dt2 = 4u2 > 0
Therefore, x2(i.e., x) is minimum at t = a/2u
Now 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.

14. 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

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 – bt2
Integrating 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) + k
Or k = 0
Thus, v = at – bt3/3   ……….(2)
Again, d2v/dt2 = d(a – bt2)/dt = -2bt
Now, for minimum or maximum value of v we have,
dv/dt = 0
Or a – bt2 = 0
Or t2 = a/b
Or 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.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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