Question

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the eastward component of the car’s velocity is given by vx(t)=(0.860m/s3)t2. What is the acceleration of the car when vx=12.0m/s?

Answer 1

The acceleration of the car when its velocity is 12.0m/s is 6.226m/s^2.

Step-by-step explanation:

The acceleration of the car can be found by taking the derivative of the velocity function. Since the velocity function is given by

`vx(t)=(0.860m/s^3)t^2,`

with respect to time to get the acceleration function

`ax(t) = 2(0.860m/s^3)t.`

when vx = 12.0m/s, we substitute vx = 12.0m/s into the velocity function and solve for t:

12.0m/s = (0.860m/s^3)t^2.

Dividing both sides by 0.860m/s^3 gives us t^2 = 13.95, and taking the square root of both sides gives us t = ±3.73s.

Since time cannot be negative, we discard the negative value and conclude that the car takes approximately 3.73 seconds to reach a velocity of 12.0m/s. Now we can substitute this value of t into the acceleration function to find the acceleration at this time: ax(3.73s) = 2(0.860m/s^3)(3.73s) = 6.226m/s^2.

Answer 2

ax = 6.43m/s²

Step-by-step explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.