33rd Putnam 1972

Problem B2

A particle moves in a straight line with monotonically decreasing acceleration. It starts from rest and has velocity v a distance d from the start. What is the maximum time it could have taken to travel the distance d?



Answer: 2d/v.

Plot velocity u(t) against time. We have u(T) = v. The area under the curve between t = 0 and t = T is the distance d. But since the acceleration is monotonically decreasing, the curve is concave and hence the area under it is at least the area of the triangle formed by joining the origin to the point t = T, u = v (other vertices t = 0, u = 0 and t = T, u = 0). Hence d ≥ 1/2 vT, so T ≤ 2d/v. This is achieved by a particle moving with constant acceleration.



33rd Putnam 1972

© John Scholes
27 Jan 2001