Question

The position x of a particle varies with time (t) as x = At² – Bt³. The acceleration at time t of the particle will be equal to zero. What is the value of t?
(a) $\frac{2A}{3B}$
(b) $\frac{A}{B}$
(c) $\frac{A}{3B}$
(d) zero

Answer 1

x = At² - Bt³

Now, velocity is dx/dt, and acceleration is dv/dt, so acceleration is d²x/dt².

Hence, 2nd derivative of x wrt. time, acceleration:

a = 2A - 6Bt

For a to be 0,

2A - 6Bt = 0

Or, t = A/3B (c).