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?
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?
Answers
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
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
Thanks
Please
Please help me
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