Question

A car is traveling at a 20.0 m/s for 7.00 s and then suddenly comes to a stop over a 3 s period.

Answer 1

Answer: A) Deceleration of the car is -6.6667 m/s² while it came to stop.

B) The total distance the car travels is 200 meter during the 10 s period.

Explanation:

Given Data

Initial velocity of the car ($v_{i}$) =   20.0 m/s

Final velocity of the car ($v_{f}$) = 0 m/s

Time (in motion) =7.00 s

Time (in rest) =3 s

To find - A) car's deceleration while it came to a stop

              B) the total distance the car travels in 10 s

A) The formula to find the deceleration is

Deceleration = (( final velocity - initial velocity ) ÷ Time)        (m/s²)

Deceleration = (($v_{f}$) - ($v_{i}$)) ÷ time     (m/s²)

Deceleration =  ( 0.0 - 20 ) ÷ 3      (m/s²)

Deceleration =   (- 20) ÷ 3  (m/s²)

Deceleration   =  - 6.6667 m/s²

(NOTE : Deceleration is the opposite of acceleration so the final result must have the negative sign)

The car's deceleration is  - 6.6667 m/s² while it came to a stop

B) The formula to find the distance traveled by the car is  

Distance traveled by the car is equals to the product of the speed and time

Distance = Speed × Time  (meter)

Distance = 20.0 × 10

Distance = 200 meters

The total distance the car travels during the period of 10 s is 200 meters