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

Answer:

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

(b). The distance is 43.35 m.

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

$v_{avg}=\dfrac{v_{f}-v_{i}}{2}$

Put the value into the formula

$v_{avg}=\dfrac{0-(-25.5)}{2}$

$v_{avg}=12.75\ m/s$

(b). We need to calculate the acceleration

Using equation of motion

$v_{f}=v_{i}+at$

$a =\dfrac{v_{f}-v_{i}}{t}$

$a=\dfrac{-25.5-0}{3.4}$

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

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=25.5\times3.4+\dfrac{1}{2}\times(-7.5)\times(3.4)^2$

$s=43.35\ m$

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

(b). The distance is 43.35 m.