Question

The velocity function of a car is given by v(t) = -3t^2 + 18t + 9 m/s. Find the acceleration of the car three seconds before it comes to a stop.


A. a = -6.46 m/s²

B. a = -2.76 m/s²

C. a = -0.46 m/s²

D. a = 2.76 m/s²

Answer 1

The acceleration of the car three seconds before it comes to a stop is 0m/s²

Given the velocity function expressed as v(t) = -3t^2 + 18t + 9

Acceleration is the change in velocity of a body with respect to time.

a(t) = dv/dt

Differentiating the velocity with respect to time will give;

dv/dt = -6t + 18

a = -6t + 18

a = -3(6) + 18

a = -18 + 18

a = 0m/s²

Hence the acceleration of the car three seconds before it comes to a stop is 0m/s²

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