Question

The distance x metres described by a car in time t seconds is given by: $x = 3t^3 − 2 t^2 + 4t - 1$. Determine the velocity and acceleration when $t = 1.5 s$

Answer 1

x = 3t³ - 2t² + 4t - 1
v = dx/dt = 9t² - 4t + 4
put t = 3/2
v = 81/4 - 6 + 4 =81/4 - 2
v = 73/4 m/sec.
a = dv/dt = 18t - 4
put t = 3/2
a = 27 - 4 = 23 m/sec².