### 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.

**Solution**

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