Question

Motion of a particle is given by equation s=(3t^3+7t^2+14t+8)m find the value of acceleration of a particle at t=1sec.

Answer 1

$v = \frac{dx}{dx} = 9 {t}^{2} + 14 t + 14$

$a = \frac{dv}{dt} = 18t + 14$

at t=1s

$a = 32m {s}^{ - 2}$