v = Vo + a t
so
v = 55 - 11 t
when is v = 0 ?
0 = 55 - 11 t
so t = 5 seconds to stop
d = Vo t + (1/2) a t^2
d = 55 (5) - (1/2) 11 (25)
= 275 - 137.5
=137.5 meters
Now do that again for Vo = 110 m/s
A race car can be slowed with a constant acceleration of -11 m/s^2
a. If the car is going 55 m/s, how many meters will it travel before it stops?
b. How many meters will it take to stop a car going twice as fast?
3 answers
Ok, thanks but i don't understand b. I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.
No, do v = Vo + a t again with
0 = 110 - 11 t
then use that t in
d = 110 t - (1/2)11 t^2
0 = 110 - 11 t
then use that t in
d = 110 t - (1/2)11 t^2