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?
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
To solve this problem, we can use kinematic equations. The kinematic equation that relates displacement (d), initial velocity (vi), final velocity (vf), and acceleration (a) is:
vf^2 = vi^2 + 2ad
This equation allows us to find the displacement when decelerating from an initial velocity down to zero.
a) If the car is going 55 m/s, we can plug in the following values:
vi = 55 m/s
vf = 0 m/s
a = -11 m/s^2
Rearranging the equation, we have:
0 = 55^2 + 2(-11)d
Simplifying,
0 = 3025 - 22d
Moving the terms around,
22d = 3025
Dividing both sides by 22,
d = 137.5 m
Therefore, the car will travel 137.5 meters before it stops.
b) If the car is going twice as fast, its initial velocity will be 2 * 55 = 110 m/s. We can use the same equation:
vi = 110 m/s
vf = 0 m/s
a = -11 m/s^2
0 = 110^2 + 2(-11)d
Simplifying,
0 = 12100 - 22d
Moving the terms around,
22d = 12100
Dividing both sides by 22,
d = 550 m
Therefore, it will take 550 meters to stop a car going twice as fast.
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