Distance traveled derivative

Question

The distance S covered by a car after t sec is given by: s=-t³ + 8t² +20t (0≤t≤6)

 

Find the general expression for the car's acceleration at any time t (0≤t≤6). Show that the car is decelerating after 2 2/3 sec. 

Answer 1

a = d²s/dt²

 

ds/dt = -3t² + 16t + 20

 

a(t) = d²s/dt² = -6t + 16

 

At 8/3 sec,

 

a(8/3) = -6(8/3) + 16 = 0

 

a > 0 when t < 8.3 (positive acceleration)

a = 0 when t = 8/3 (instantaneously zero acceleration)

a < 0 when t > 8/3 (negative acceleration, i.e. deceleration)