Question

In a test of a prototype car, the driver starts the car from rest at t=0, accelerates, and then applies the brakes. Engineers measuring the position of the car find that from t=0 to t=18s. The position is approximated by s=5t^2 1/3t^3 -1/50t^4 ft.
a.) what is the maximum velocity, and at what time does it occur?
b.) What is the maximum acceleration, and at what time does it occur?

Answer 1

I'll assume you meant

s = 5t^2 + 1/3t^3 -1/50t^4
so,
v = 10t + t^2 - 2/25 t^3
max v occurs when v'=0, so
a = 10 + 2t - 6/25 t^2

So, find t when a=0, and then figure v at that value of t.

max acceleration is when a'=0
a' = 2 - 12/25 t

Answer 2

Thanks

Answer 3

Please

Answer 4

Please help me