This set of Mathematics Problems for Engineering Entrance Exams focuses on “Calculus Application – Velocity – 2”.

1. 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 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

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)

And f = dv/dt = d(b + ct)/dt = c ……….(3)

Clearly, when c = 0, then f = 0 that is, acceleration of the particle is zero.

Hence in this case the particle moves with an uniform velocity.

2. A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t^{2} – 4t + 5 cm/second. What will be the distance travelled by the particle during first 3 seconds after the start?

a) 21 cm

b) 22 cm

c) 23 cm

d) 24 cm

View Answer

Explanation: Let, x be the distance travelled by the particle in time t seconds.

Then, v = dx/dt = 3t

^{2}– 4t + 5

Or ∫dx = ∫ (3t

^{2}– 4t + 5)dt

So, on integrating the above equation, we get,

x = t

^{3}– 2t

^{2}+ 5t + c where, c is a constant. ……….(1)

Therefore, the distance travelled by the particle at the end of 3 seconds,

= [x]

_{t = 3}– [x]

_{t = 0}

= (3

^{3}– 2*3

^{2}+ 5*3 + c) – c [using (1)]

= 24 cm.

3. A particle moving in a straight line traverses a distance x in time t. If t = x^{2}/2 + x, then which one is correct?

a) The retardation of the particle is the cube of its velocity

b) The acceleration of the particle is the cube of its velocity

c) The retardation of the particle is the square of its velocity

d) The acceleration of the particle is the square of its velocity

View Answer

Explanation: We have, t = x

^{2}/2 + x

Therefore, dt/dx = 2x/2 + 1 = x + 1

Thus, if v be the velocity of the particle at time t, then

v = dx/dt = 1/(dt/dx)

= 1/(x + 1) = (x + 1)

^{-1}

Thus dv/dt = d((x + 1)

^{-1})/dt

= (-1)(x + 1)

^{-2}d(x + 1)/dt

= -1/(x + 1)

^{2}* dx/dt

As, 1/(x + 1) = dx/dt,

So, -(dx/dt)

^{2}(dx/dt)

Or dv/dt = -v

^{2}*v [as, dx/dt = v]

= -v

^{3}

We know, dv/dt = acceleration of a particle.

As, dv/dt is negative, so there is a retardation of the particle.

Thus, the retardation of the particle = -dv/dt = v

^{3}= cube of the particle.

4. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?

a) 30 cm

b) 31 cm

c) 32 cm

d) 33 cm

View Answer

Explanation: We assume that the particle moves with uniform acceleration 2f m/sec.

Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.

Let, v be the velocity of the particle at time t seconds, then,

So, dv/dt = 2f

Or ∫dv = ∫2f dt

Or v = 2ft + b ……….(1)

Or dx/dt = 2ft + b

Or ∫dx = 2f∫tdt + ∫b dt

Or x = ft

^{2}+ bt + a ……….(2)

Where, a and b are constants of integration.

Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.

Putting these values in (2) we get,

4f + 2b + a = 21 ……….(3)

16f + 4b + a = 43 ……….(4)

49f + 7b + a = 91 ……….(5)

Solving (3), (4) and (5) we get,

a = 7, b = 5 and f = 1

Therefore, from (2) we get,

x = t

^{2}+ 5t + 7

Therefore, the distance described by the particle in 3 seconds,

= [x]

_{t = 3}= (3

^{2}+ 5*3 + 7)m = 31m

5. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?

a) 11m/sec

b) 31 cm/sec

c) 21m/sec

d) 41m/sec

View Answer

Explanation: We assume that the particle moves with uniform acceleration 2f m/sec.

Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.

Let, v be the velocity of the particle at time t seconds, then,

So, dv/dt = 2f

Or ∫dv = ∫2f dt

Or v = 2ft + b ……….(1)

Or dx/dt = 2ft + b

Or ∫dx = 2f∫tdt + ∫b dt

Or x = ft

^{2}+ bt + a ……….(2)

Where, a and b are constants of integration.

Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.

Putting these values in (2) we get,

4f + 2b + a = 21 ……….(3)

16f + 4b + a = 43 ……….(4)

49f + 7b + a = 91 ……….(5)

Solving (3), (4) and (5) we get,

a = 7, b = 5 and f = 1

Therefore, from (2) we get,

x = t

^{2}+ 5t + 7

Putting t = 3, f = 1 and b = 5 in (1),

We get, the velocity of the particle in 3 seconds,

= [v]

_{t = 3}= (2*1*3 + 5)m/sec = 11m/sec.

6. A particle moves in a horizontal straight line under retardation kv^{3}, 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^{2} = 1/2u^{2} + 2kt

b) 1/v^{2} = 1/2u^{2} – 2kt

c) 1/v^{2} = 1/u^{2} – 2kt

d) 1/v^{2} = 1/u^{2} + 2kt

View Answer

Explanation: Since the particle is moving in a straight line under a retardation kv

^{3}, hence, we have,

dv/dt = -kv

^{3}……….(1)

Or dv/v

^{3}= -k dt

Or ∫v

^{-3}dv = -k∫dt

Or v

^{-3+1}/(-3 + 1) = -kt – c [c = constant of integration]

Or 1/2v

^{2}= kt + c ……….(2)

Given, u = v when, t = 0; hence, from (2) we get,

1/2u

^{2}= c

Thus, putting c = 1/2u

^{2}in (2) we get,

1/2v

^{2}= kt + 1/2u

^{2}

Or 1/v

^{2}= 1/u

^{2}+ 2kt

7. 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 = ax^{2} + bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?

a) 1/v^{2} + 1/u^{2} = 4at

b) 1/v^{2} + 1/u^{2} = -4at

c) 1/v^{2} – 1/u^{2} = 4at

d) 1/v^{2} – 1/u^{2} = -4at

View Answer

Explanation: We have, t = ax

^{2}+ bx + c ……….(1)

Differentiating both sides of (1) with respect to x we get,

dt/dx = d(ax

^{2}+ 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)

Initially, when t = 0 and v = u, let x = x

_{0}; hence, from (1) we get,

ax

_{0}

^{2}+ bx

_{0}+ c = 0

Or ax

_{0}

^{2}+ bx

_{0}= -c ……….(3)

And from (2) we get, u = 1/(2ax

_{0}+ b)

Thus, 1/v

^{2}– 1/u

^{2}= (2ax + b)

^{2}– (2ax

_{0}+ b)

^{2}

= 4a

^{2}x

^{2}+ 4abx – 4a

^{2}x

_{0}

^{2}– 4abx

_{0}

= 4a

^{2}x

^{2}+ 4abx – 4a(ax

_{0}

^{2}– bx

_{0})

= 4a

^{2}x

^{2}+ 4abx – 4a(-c) [using (3)]

= 4a(ax

^{2}+ bx + c)

Or 1/v

^{2}– 1/u

^{2}= 4at

8. The distance s of a particle moving along a straight line from a fixed-pointO on the line at time t seconds after start is given by x = (t – 1)^{2}(t – 2)^{2}. What will be the distance of the particle from O when its velocity is zero?

a) 4/27 units

b) 4/23 units

c) 4/25 units

d) 4/35 units

View Answer

Explanation: Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,

v = ds/dt = d[(t – 1)

^{2}(t – 2)

^{2}]/dt

Or v = (t – 2)(3t – 4)

Clearly, v = 0, when (t – 2)(3t – 4) = 0

That is, when t = 2

Or 3t – 4 = 0 i.e., t = 4/3

Now, s = (t – 1)(t – 2)

^{2}

Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)

^{2}= 4/27

And when t = 2, then s =(2 – 1)(2 – 2)

^{2}= 0

Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.

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