Help with a Physics question.?

Question

I tried a hundred times and cant figure out how to do this problem without initial velocity. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration -5.40 m/s2 for 4.30 s, making straight skid marks 62.5 m long ending at the tree. With what speed does the car then strike the tree?

Answer 1

The first working formula is

S = Vo(T) + (1/2)(a)T^2

where

S = distance travelled = 62.5 m (given)

Vo = initial velocity of car

T = travel time of car = 4.3 sec (given)

a = acceleration = -5.4 m/sec^2

Substituting values,

42.5 = Vo(4.3) + (1/2)(-5.4)(4.3)^2

4.3 Vo = 42.5 + (1/2)(5.4)(4.3)^2

Vo = (1/4.3)(42.5 + (1/2)(5.4)(4.3)^2)

Vo = 26.15 m/sec.

The second formula to be used is

Vf^ - Vo = aT

where

Vf = velocity at which car will strike the tree

and all the other terms have been previously defined.

Substituting values,

Vf - 26.15 = (-5.4)(4.3)

Vf = 26.15 - 5.4(4.3)

Vf = 2.93 m/sec

Answer 2

s= 62.5m (displacement)

t = 4.3s (time)

a = -5.4ms^-2 (acceleration)

v=? (final velocity)

u=? (initial velocity)

you are right you need initial velocity which can be found out using:

s = ut + 0.5at^2

62.5 = 4.3u - 2.7 x 4.3^2

62.5 +2.7 x 4.3^2 = 4.3u

u = (62.5 + 2.7 x 4.3^2)/4.3

knowing u, sub this into v^2 = u^2 + 2as to find final velocity