On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only 5.00 m/s2 . Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Explanation:

a ) Let the distance required in former case be d₁ .

initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 7 x d₁

d₁ = 64.28 m

b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 5 x d₂

d₂ = 90  m

c)

t = .5 s

s₁  = ut - .5 at²

= 30 x .5 - .5 x 7 x .5²

= 15 - .875

= 14.125 m

t = .5 s

s₂  = ut - .5 at²

= 30 x .5 - .5 x 5 x .5²

= 15 - .625

= 14.375  m