Question

A car moving with an initial velocity of 45 ms¹ is brought to a stop in 5 s. Find:

a) the deceleration of the car

b)the distance needed to stop the car​

Answer 1

a) -9
`m/s^2`

b) 112.5 meters

Step-by-step explanation:

Here is a formula for acceleration.

`a=(V_(f) -V_(o) )/(t)`

We are given

`V_o=45` m/s

`V_f=0` m/s

`t=5` s

Lets solve for
`a`.

`a=(0-45)/(5)`

`a=(-45)/(5)`

`a=-9`
`m/s^2`

Here is a formula for displacment.

`x=V_o t+(at^2)/(2)`

`V_o=45` m/s

`t=5` s

`a=-9`
`m/s^2`

Lets solve for
`x`.

`x=45*5+(-9*5^2)/(2)`

`x=225+(-9*25)/(2)`

`x=225+(-225)/(2)`

`x=225+-112.5`

`x=225-112.5`

`x=112.5`