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?
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To find the maximum velocity and the time at which it occurs, we need to find the derivative of the position function with respect to time (t). Once we have the derivative, we can set it to zero and solve for t to find the time at which the maximum velocity occurs.
a.) Maximum velocity and the time it occurs:
1. Start by finding the derivative of the position function with respect to time (t):
s(t) = 5t^2 + (1/3)t^3 - (1/50)t^4
Taking the derivative:
s'(t) = d/dt (5t^2 + (1/3)t^3 - (1/50)t^4)
To find the derivative of each term, we can use the power rule:
s'(t) = 10t + t^2 - (2/25)t^3
2. Set the derivative equal to zero and solve for t:
10t + t^2 - (2/25)t^3 = 0
To solve this equation, you can use numerical methods or approximation techniques such as graphing or using a calculator.
b.) Maximum acceleration and the time it occurs:
To find the maximum acceleration, we need to find the derivative of the velocity function with respect to time. The velocity function is the derivative of the position function.
1. Start by finding the derivative of the velocity function with respect to time (t):
v(t) = s'(t) = 10t + t^2 - (2/25)t^3
Taking the derivative:
v'(t) = d/dt (10t + t^2 - (2/25)t^3)
To find the derivative of each term, we can again use the power rule:
v'(t) = 10 + 2t - (6/25)t^2
2. Set the derivative equal to zero and solve for t:
10 + 2t - (6/25)t^2 = 0
Again, you can use numerical methods or approximation techniques to solve this equation and find the time at which the maximum acceleration occurs.
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