A car is initially traveling at 17 m/s when the driver steps down hard on the gas pedal

resulting in an acceleration of 3.0 m/s2. How fast will the car be moving 4.0 seconds later? v=29

m/s

Question

Hodges

Physics

12 November, 14:27

A car is initially traveling at 17 m/s when the driver steps down hard on the gas pedal

resulting in an acceleration of 3.0 m/s2. How fast will the car be moving 4.0 seconds later? v=29

m/s

+2

Answers (1)

Know the Answer?

Skyla Rose · Answer 1

29 m/s

Explanation:

Acceleration is the rate of change of velocity, i. e, the change in velocity per unit time.

The driver made the car to accelerate at 3 m/s^2. That means than in every second the velocity of the car increased by 3 m/s. Therefore in four seconds its velocity would increase by 3m/s x 4 = 12m/s. Starting with an initial velocity of 17 m/s the final velocity would be 17 m/s + 12 m/s = 29 m/s.

A car is initially traveling at 17 m/s when the driver steps down hard on the gas pedalresulting in an acceleration of 3.0 m/s2. How fast will the car be moving 4.0 seconds later? v=29m/s

A car is initially traveling at 17 m/s when the driver steps down hard on the gas pedal

resulting in an acceleration of 3.0 m/s2. How fast will the car be moving 4.0 seconds later? v=29

m/s