## Forgot to use ^, here is the Q again. Thank you

I had help earlier with the below, but I had to make a couple of changes to make sure all numbers were in the right SI units. Would someone be kind enough to check over my calculation as I think something is wrong. I am having problems with converting SI units and so this might be the source of the mistake.

The question was 'A truck is moving at 88 km h^-1 and breaks to 52km h ^-1. What is the distance over which the breaks are used if the deceleration was 2.82 m s^-2.'

using the following equation:

Ek = 1/2mv^2

W = Fd

F = ma

Calculation

Ek = ½ mv^2

At 88 km h1 (88 km x 1000 x h / 3600 ^−1 = 2.82 x 101m s^-1) the car has kinetic energy of:

Ek = ½m(2.44 x 101m s^-1)^2 Joules

At 52 km
h^1 (52 km x 1000 x h / 3600^1 = 1.44 x 101m s^-1)the car has kinetic energy:

Ek = ½m(1.44 x 101m s^-1)^2 Joules

The work of the brakes is difference between the two =

½m(2.44 x 101m s^-1)^2 -½m(1.44 x 101m s^-1)^2 Joules

= (
½)m(2.44 x 101 m s – 1.44 x 101 m s^-1 )^2 Joules

The car decelerates at 2.82 m s^−2 = acceleration - 2.82 m s^−2

F=ma
= F = -m(2.82 m s^
2)

Work = force x distance (W=Fd)
or Fd = W
Fd / F = W/F
d = W / F

d = W/F

d = (½)m(2.44 x 101 m s – 1.44 x 101 m s^-1) 2 Joules x -m(2.82 m s^2)

(
½)m x – m = 0 (and so cancels itself out)

d = (2.44 x 101 m s – 1.44 x 101 m s^-1) 2 Joules x (2.82 m s^2)

d = 1.00 x 102 x 7.95 m

d = 7.95 x 102 m