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 = ˝ mv2
At 88 km h −1 (88 km x 1000 x h / 3600 −1 = 2.82 x 101m s-1) the car has kinetic energy of:
Ek = ˝m(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 = ˝m(1.44 x 101m s-1)2 Joules
The work of the brakes is difference between the two =
˝m(2.44 x 101m s-1)2 -˝m(1.44 x 101m s-1)2 Joules
= (˝)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 s2)
Work = force x distance (W=Fd)
or Fd = W
Fd / F = W/F
d = W / F
d = W/F
d = (˝)m(2.44 x 101 m s – 1.44 x 101 m s-1) 2 Joules x -m(2.82 m s2)
(˝)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 s2)
d = 1.00 x 102 x 7.95 m
d = 7.95 x 102 m