This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Velocity – 1”.

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 – 15t^{2} + 4t^{3}. What is the velocity of the particle after 3 seconds?

a) 10 cm/sec

b) 20 cm/sec

c) 30 cm/sec

d) 40 cm/sec

View Answer

Explanation: We have, s = 12t – 15t

^{2}+ 4t

^{3}……….(1)

Differentiating both side of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t

^{2}

So, velocity of the particle after 3 seconds is,

[ds/dt]

_{t = 3}= 12 – 30(3) + 12(3)

^{2}

= 30 cm/sec.

2. 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 – 15t^{2} + 4t^{3}. What will be the distance between the two positions of the particle at two times, when the velocity is instantaneously 0?

a) 27/4 cm

b) 29/4 cm

c) 27/2 cm

d) 29/2 cm

View Answer

Explanation: We have, s = 12t – 15t

^{2}+ 4t

^{3}……….(1)

Differentiating both side of (1) with respect to t we get,

(ds/dt) = 12 – 30t + 12t

^{2}

Clearly, the velocity is instantaneously zero, when

(ds/dt) = 12 – 30t + 12t

^{2}= 0

Or 12 – 30t + 12t

^{2}= 0

Or (2t – 1)(t – 2) = 0

Thus, t = 2 or t = ½

Putting the value t = 2 and t = ½ in (1),

We get, when t = 2 then s = (s1) = 12(2) – 15(2)

^{2}+ 4(2)

^{3}= -4.

When t = ½ we get, s = (s2) = 12(1/2) – 15(1/2)

^{2}+ 4(1/2)

^{3}= 11/4.

Thus, the distance between the two positions of the particle at two times, when the velocity is instantaneously 0 = s2 – s1

= 11/4 – (-4)

= 27/4 cm.

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 = 2t^{3} – 12t + 11. What is the velocity of the particle at the end of 2 seconds?

a) 10 cm/sec

b) 12 cm/sec

c) 14 cm/sec

d) 16 cm/sec

View Answer

Explanation: We have, x = 2t

^{3}– 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(2t

^{3}– 12t + 11)/dt

= 6t

^{2}– 12 ……….(2)

Putting the value of t = 2 in (2),

Therefore, the displacement of the particle at the end of 2 seconds,

6t

^{2}– 12 = 6(2)

^{2}– 12

= 12 cm/sec.

4. 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 = 2t^{3} – 12t + 11. What is the displacement of the particle at the end of 2 seconds?

a) 1 cm

b) 2 cm

c) 3 cm

d) 4 cm

View Answer

Explanation: We have, x = 2t

^{3}– 12t + 11 ……….(1)

Putting t = 2 in (1), we get,

Displacement = x = 2t

^{3}– 12t + 11

= 2(2)

^{3}– 12(2) + 11

= 3 cm.

5. 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 = t^{3} – t^{2} – 5t. What will be the distance traversed before it comes to rest?

a) -173/27

b) 173/27

c) -175/27

d) 175/27

View Answer

Explanation: We have, x = t

^{3}– t

^{2}– 5t ……….(1)

When x = 28, then from (1) we get,

t

^{3}– t

^{2}– 5t = 28

Or t

^{3}– t

^{2}– 5t – 28 = 0

Or (t – 4)(t

^{2}+ 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(t

^{3}– t

^{2}– 5t)/dt

= 3t

^{2}– 2t – 5

And f = dv/dt = d(3t

^{2}– 2t – 5)/dt

= 6t – 2

Again the particle comes to rest when v = 0

Or 3t

^{2}– 2t – 5 = 0

Or (3t – 5)(t + 1) = 0

Or t = 5/3, -1

As, t > 0, so, t = 5/3

Therefore, the distance traversed by the particle before it comes to rest

= [(5/3)

^{3}– (5/3)

^{2}– 5(5/3)] m [putting t = 5/3 in (1)]

= -175/27

6. “It is impossible for a particle to move in a straight line so that its velocity varies at the distance from the commencement of motion”. Which one is correct for the given statement?

a) The above statement is valid

b) The above statement is not valid

c) Data inadequate

d) Answer does not exist

View Answer

Explanation: Let a particle be moving along the straight-line OX and P be its position at time t, when OP = x. Clearly, the velocity and acceleration of the particle at P are dx/dt and d

^{2}x/dt

^{2}respectively.

If possible, let us assume,

dx/dt α x

dx/dt = kx, where k is a constant variation.

Now, d

^{2}x/dt

^{2}= d(kx)/dt = k(dx/dt)

= k

^{2}x

At the starting point O, we have x = 0 and we see that dx/dt = 0 and d

^{2}x/dt

^{2}= 0, when x = 0 that is, both velocity and acceleration of the particle are zero at x = 0 and the particle remains at rest at O.

Therefore, it is impossible for a particle to have any motion under the given condition.

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 + ct^{2} (a, b > 0). What is the meaning of the constant a?

a) Initial position

b) Final position

c) Mid position

d) Any arbitrary position

View Answer

Explanation: We have, x = a + bt + ct

^{2}……….(1)

Now from (1) we get x = a, when t = 0.

This means that the distance of the particle from O at time t = 0 is a ft.

Thus, a represents the initial position of the particle.

8. 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 + ct^{2} (a, b > 0). What is the meaning of the constant b?

a) Final velocity

b) Initial velocity

c) Mid velocity

d) Arbitrary velocity

View Answer

Explanation: We have, x = a + bt + ct

^{2}……….(1)

Let, v and f be the velocity and acceleration of a particle at time t seconds.

Then, v = dx/dt = d(a + bt + ct

^{2})/dt = b + ct ……….(2)

Now from (2) we get v = b, when t = 0.

Hence b represents the initial velocity of the particle.

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