Question

While driving to school this morning a deer ran out in front of my car. When I first saw the deer, it was 35 m away. It took 0.70 seconds before I hit my brakes and then I started slowing with an acceleration of -10 m / (s ^ 2) I was able to stop 3 meters before hitting the deer. How fast was going before I saw the deer?

Answer 1

Step-by-step explanation:

vo = initial speed in m/s

in .7 seconds you will cover

.7 * vo

then your stopping distance is :

35 -.7vo - 3 = 32 - . 7vo meters

Using standard formulas

vf = vo - a t

0 = vo - 10 t shows vo = 10t where t is the time the brakes were on

d = vo t + 1/2 a t^2 a = -10 m/s^2

32 - .7 (10t) = 10 t * t + 1/2 (-10)t^2

simplify to

5t^2 + 7t - 32 = 0 Use Quadratic formula to find t = 1.92488 seconds

then

vf = vo + at

0 = vo + (-10)1.92488 vo = 19.25 m/ s initial velocity

CHECK:

35 = 3 + .7 (19.25) + 19.25 (1.92488) + 1/2 (-10) (1.92488)^2 ?

35 = 35 CHECK !