A particle moves in one dimension. Its distance x from the origin at time t is at + bt2 + ct3. Find an expression for the particle's acceleration in terms of a, b, c and its speed v.
Differentiating, v = 3ct2 + 2bt + a (*), and hence the acceleration f = 6ct + 2b. So t = (f - 2b)/6c. Substituting in (*) gives v = (f - 2b)2/12c + b(f - 2b)/3c + a. Hence 12cv = f2 - 4b2 + 12ac. So f = 2√(b2 + 3cv - 3ac).
6th Putnam 1946
© John Scholes
16 Dec 1999