A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given by x(t) = bt^2 - ct^3, where b = 3 m/s^2 and c = 0.110 m/s^3.
a) Calculate the average velocity of the car for the time interval t = 0 to t = 10s.
b) Calculate the instantaneous velocity of the car at t = 0, t = 5.0s, and t = 10.0s
c) How long after starting from rest is the car again at rest?
So I solved part A - 19 m/s just plugging t = 10 into the given formula.
I'm not sure how to solve part b, especially since I thought instantaneous velocity at t = 0 should be 0 given that the car starts at rest. However, my book's answer key says that the velocity at this instant is not zero. What am I missing for calculating instantaneous velocity?
You are missing nothing. Taking the derivative of position with respect to time
v=dx/dt= 2bt - 3ct^2, and at t=0, v=0 The answer key is wrong.
To calculate the instantaneous velocity at a specific time, we need to find the derivative of the position function with respect to time, which gives us the rate of change of position with respect to time. In this case, the position function is x(t) = bt^2 - ct^3.
Taking the derivative of x(t) with respect to t, we get:
v(t) = d(x(t))/dt = d(bt^2 - ct^3)/dt
To find v(t), we differentiate each term of the equation separately:
v(t) = d(bt^2)/dt - d(ct^3)/dt
Using the power rule of derivatives, where the derivative of t^n is n*t^(n-1):
v(t) = 2bt - 3ct^2
Now, we can find the instantaneous velocity at different points in time.
a) At t = 0:
Substituting t = 0 into the velocity function v(t), we get:
v(0) = 2b(0) - 3c(0)^2
= 0
Therefore, the instantaneous velocity at t = 0 is indeed 0, which means the car is at rest at that moment. If the answer key claims otherwise, there may be an error.
b) At t = 5.0s:
Substituting t = 5.0 into the velocity function v(t), we get:
v(5.0) = 2b(5.0) - 3c(5.0)^2
Substitute the values for b and c:
v(5.0) = 2(3)(5.0) - 3(0.110)(5.0)^2
Calculating this expression will give you the instantaneous velocity at t = 5.0s.
c) At t = 10.0s:
Substituting t = 10.0 into the velocity function v(t), we get:
v(10.0) = 2b(10.0) - 3c(10.0)^2
Substitute the values for b and c, and calculate to find the instantaneous velocity at t = 10.0s.
To solve part c, we need to find the time at which the car is again at rest. This is equivalent to finding the time at which the velocity, v(t), is equal to zero.
Setting v(t) = 0, we have:
0 = 2bt - 3ct^2
Now, we can solve this equation for t to find the time at which the car comes to rest. This can be done by factoring, quadratic formula, or numerical methods like Newton's method.