Question

If the velocity of car at time t(sec) is directly proportional to the square of its velocity at time (t-1)(sec). Then find the ratio of acceleration at t=10sec to 9sec if proportionality constant is k=10 sec/mt and velocity at t=9sec is 10 mt/sec.

(a) 100

(b) 200

(c) 150

(d) 250

I got this question in an internship interview.

My question is taken from Implicit Differentiation in chapter Partial Differentiation of Engineering Mathematics

Answer 1

Right option is (b) 200

Easy explanation: Given,v(t)=kv^2 (t-1)

Differentiating w.r.t time we get

^dv(t)⁄dt = 2kv(t-1) ^dv(t – 1)⁄dt

a(t) = 2*10*10 a(t-1)

^a(t)⁄a(t – 1) = 200.