Show that the equation for the stopping distance of a car is d = vt - v^2/2a?

Question

Show that the equation for the stopping distance of a car is d = vt - v^2/2a

v = initial velocity

d = stopping distance

t = reaction time

a = acceleration (negative)

How do u prove this???!

Answer 1

given that

1) d = v(avg)t

2) v(avg) = (vi+vf)/2   where vi= initial vel, vf=final vel

3) a = (vf-vi)/t

sub eq 2 into eq 1

d=(vi+vf)t/2

expand brackets

d=vit/2 +vft/2    (lets call this eq. 4)

solve eq 3 for vf

vf= vi+at

and sub this into eq 4

d=vit/2 + (vi+at)t/2

expand brackets

d=vit/2 +vit/2 + at^2/2

simplify

d= vit+1/2at^2

hmmm.. this isn't what you have... are you sure you don't have a typo in your formula?

Answer 2

if driver sees the object at t=0

and applies breaks at t=t

then reaction time = t

---------------------------

at t=0, velocity =v

kept moving with v for time = t

s1 = vt = distance covered

decelerated with (a) from t=t and stopped after covering (s2)

vf^2 = v^2 + 2 a [s2]

when stops vf=0

s2 = - v^2/2a

d = stopping distance = s1+s2 = vt - v^2/2a

Answer 3

Actually the equation is:

d = vt + v²/2μa

where μ is the effective coefficient of friction between the tires and the road.

Use the conservation of energy. The car's initial kinetic energy is ½mv², and it's final kinetic energy is 0. The energy to stop the car is μ*m*a*d.

μ*m*a*d = ½mv²

d = v²/2μa = distance traveled after breaks applied

Total distance traveled:

d = vt + v²/2μa

I don't know why you have "- v²/2a." Although the car is decelerating, it's still traveling in the same direction.

EDIT: Correction. Ignore the "μ" in the equation. The effective coefficient of friction is already factored into the deceleration.