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) = (0.890m/s3) t2.

Part A

What is the acceleration of the car whenvx = 14.1 m/s?

Express your answer with the appropriate units.

ax=

a_x (3.98) = 7.0844 m/s^2

Explanation:

Given:

- The eastward component of car's velocity is given as follows:

v_x (t) = 0.890*t^2

For, 0 s < t < 5 s

Find:

What is the acceleration of the car when v_x = 14.1 m/s?

Solution:

- We will first compute the time t at which the velocity v_x component equals 14.1 m / s:

14.1 = 0.890*t^2

t = sqrt (15.8427)

t = 3.98 s

- Now use the relation of v_x and derivative to obtain a_x acceleration in eastward direction:

a_x (t) = dv_x / dt

a_x (t) = 1.78*t

- Evaluate at t = 3.98 s:

a_x (3.98) = 1.783.98

a_x (3.98) = 7.0844 m/s^2