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 υx (t) = (0.980m/s3) t2.

What is the acceleration of the car when υx = 14.5 m/s?

Question

Romeo Zuniga

Physics

21 July, 14:58

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 υx (t) = (0.980m/s3) t2.

What is the acceleration of the car when υx = 14.5 m/s?

+3

Answers (1)

Know the Answer?

Craig Mann · Answer 1

7.54 m/s²

Explanation:

uₓ (t) = (0.980 m/s³) t²

Acceleration is the derivative of velocity with respect to time.

aₓ (t) = 2 (0.980 m/s³) t

aₓ (t) = (1.96 m/s³) t

When uₓ = 14.5 m/s, the time is:

14.5 m/s = (0.980 m/s³) t²

t = 3.85 s

Plugging into acceleration equation:

aₓ = (1.96 m/s³) (3.85 s)

aₓ = 7.54 m/s²

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 υx (t) = (0.980m/s3) t2.What is the acceleration of the car when υx = 14.5 m/s?

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 υx (t) = (0.980m/s3) t2.

What is the acceleration of the car when υx = 14.5 m/s?