Question

A car driving at an initial speed of 10.0 m/s accelerates on a straight road at 3.00 m/s^2. a) what is the speed of the car after one quarter of a mile? (1 mile =1.609km). B) the driver in part A slams on the brakes after reaching the quarter mile. If the car can decelerate at a rate of 4.50 m/s^2, what is the stopping distance of the car?

Answer 1

The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.

Step-by-step explanation:

Given that,

Initial speed = 10.0 m/s

Acceleration = 3.00 m/s^2

Distance
`d = (1)/(4)*1.609`

`d = 0.40225\ km=0.40225*10^(3)\ m`

We need to calculate the speed of the car,

Using equation of motion

`v^2-u^2=2as`

Where, u = initial velocity

v = final velocity

a = acceleration

Put the value in the equation

`v^2=(10.0)^2+2*3.00*0.40225*10^(3)`

`v^2=2513.5`

`v=50.13\ m/s`

(B). We need to calculate the stopping distance of the car,

Using equation of motion again

`v^2=u^2+2as`

Here,initial velocity = 50.13 m/s

Final velocity = 0

Acceleration = -4.50 m/s²

Put the value in the equation

`0=(50.13)^2-2*4.50* s`

`s=((50.13)^2)/(2*4.50)`

`s=279.22\ m`

Hence, The speed of the car and the stopping distance are 50.13 m/s and 279.22 m.