The motion of the car has a constant acceleration. In this case we will use the expression of the acceleration and the expression of distance for the uniformly varied movement.

The acceleration is defined as:

a = (v – v0)/t

Where:

v – v0 → Variation of speed, at time t

The speed variation, is from zero to 21 km/h in a time of 5.6 seconds.

21 km/h = (21)(1000/3600) = 5.83 m/s

Then the acceleration is:

a = (5.83 – 0)/5.6 = 1.04 m/s^2

Now, we apply the expression of distance:

d = v0t + at^2/2

d = 0 + (1.04)(5.6)^2/2

d = 16.3 m

**Then, this car travels 16.3 m in a time of 5.6 s**

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now