t=0 is x(t)= (3t^2-5t-2)(t-2)^2. Find the value of t

Question

A particle starts at time t=0 and moves along the x-axis so that its position at any time t≥0 is x(t)= (3t^2-5t-2)(t-2)^2. Find the value of t when the acceleration is zero and the particle is moving

Answer 1

Find the second derivative of the given function using the product rule and set it equal to zero. 

 

The value of t at which the second derivative equals zero is your answer. 

 

Your second derivative would be

 

6t^2 - 17t + 10 

 

Now set this equal to zero and find the value of t

 

(t -2)(6t - 5) = 0

 

 The times at which the acceleration is zero are 5/6 seconds and 2 seconds.