Question

A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the value of space described?

(a) 172 cm

(b) 176 cm

(c) 178 cm

(d) 174 cm

I have been asked this question in semester exam.

My doubt stems from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

Answer 1

The correct answer is (b) 176 cm

For explanation I would say: Let, the particle moving with a uniform acceleration of f cm/sec^2.

By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.

Thus, using the formula v = u + ft we get

34 = 10 + f*8

Or 8f = 24

Or f = 3

Therefore, the required acceleration of the particle is 3 cm/sec^2.

Thus, the space described by the particle in 8 seconds,

= [10*8 + 1/2(3)(8*8)  [using the formula s = ut +1/2(ft^2)]

= 80 + 96

= 176 cm.