Question

The driver of a car traveling at a speed of 25.5 m/s slams on the brakes and comes to a stop in 3.4 s. If we assume that the car's speed changed at a constant rate (constant net force), find the following. What was the car's average speed during this 3.4 s interval? ______ m/s

How far did the car go in this 3.4 s interval? ____ m

Answer 1

(a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.

Step-by-step explanation:

Given that,

Initial speed = -25.5 m/s

Final speed = 0

Time = 3.4 s

(a). We need to calculate the average speed

Using formula of average speed

`v_(avg)=(v_(f)-v_(i))/(2)`

Put the value into the formula

`v_(avg)=(0-(-25.5))/(2)`

`v_(avg)=12.75\ m/s`

(b). We need to calculate the acceleration

Using equation of motion

`v_(f)=v_(i)+at`

`a =(v_(f)-v_(i))/(t)`

`a=(-25.5-0)/(3.4)`

`a=-7.5\ m/s^2`

We need to calculate the distance

Using equation of motion

`s = ut+(1)/(2)at^2`

`s=25.5*3.4+(1)/(2)*(-7.5)*(3.4)^2`

`s=43.35\ m`

Hence, (a). The average speed of the car is 12.75 m/s

(b). The distance is 43.35 m.