a car traveling at 43m/s decelerates at a rate of 7.1 m/s squared. how long did it take the car to stop? what distance did the car travel while stopping?

To calculate the time it takes for the car to stop, we need to use the formula for deceleration:

v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s because the car stops)
u = initial velocity (43 m/s)
a = acceleration (deceleration in this case, -7.1 m/s^2)
s = distance traveled while stopping

Let's solve for time (t):
0^2 = 43^2 + 2(-7.1)s

Rearranging the equation, we get:
49.6s = 1849

Now we can solve for s:
s = 1849 / 49.6
s ≈ 37.28 m

Therefore, the car traveled approximately 37.28 meters while stopping.

Next, let's solve for time (t):
0^2 = 43^2 + 2(-7.1)s

Rearranging the equation, we get:
49.6s = 1849

Now we can solve for s:
s = 1849 / 49.6
s ≈ 37.28 m

Therefore, the car traveled approximately 37.28 meters while stopping.

v = Vi -7.1 t

0 = 43 - 7.1 t
t = 43/7.1 = 6.06 s

average speed during stop = 7.1 /2
=3.55 m/s

d = 3.55 * 6.06 = 21.5 m