6th Putnam 1946 Problem A6 A particle moves in one dimension. Its distance x from the origin at time t is at + bt 2 + ct 3 . Find an expression for the particle's acceleration in terms of a, b, c and its speed v.
Differentiating, v = 3ct 2 + 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 = f 2 - 4b 2 + 12ac. So f = 2√(b 2 + 3cv - 3ac). 6th Putnam 1946 © John Scholes jscholes@kalva.demon.co.uk 16 Dec 1999