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?
I know you use the equation d = Vit +1/2 at^2 but to find the distance would you just calculate d = vit only.
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
Vo = 110 m/s
3 answers
Ok i think i understand. I'm pretty sure you have to use the full equation d= vit +1/2 at^2
a) The average velocity when decelerating will be Vo/2 = 27.5 m/s. Multiply that by the time required to stop, Vo/a, to get the stopping distance X.
X = (Vo/a) * (Vo/2)
X = Vo^2/(2 a)= (55)^2/(22) = 137.5 m
Same answer you got, but I used a shortcut
X = (Vo/a) * (Vo/2)
X = Vo^2/(2 a)= (55)^2/(22) = 137.5 m
Same answer you got, but I used a shortcut
HAPPY NEW YEAR! Ok, thanks a lot.
1 answer