Question

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

I got this question in exam.

This is a very interesting question from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer 1

The correct option is (a) 11m/sec

The explanation is: 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.