A car comes to a stop for a stop sign in 4.6 seconds. If it was initially traveling at 27.5 m/s, how much space did it take for the car to stop?

Vf=Vi+at solve for a.

then
Vf^2=Vi^2 + 2ad solve for d.

what is Vi?

vi=27.5m/s

this is where get stuck

vf=27.5(a)(4.6)
vf=126.5
i don't know the rest

please help me im realy stuck

To determine the space the car took to stop, we need to calculate the distance traveled during the deceleration.

The equation to find the distance traveled during deceleration is:

distance = (initial velocity * time) + (0.5 * acceleration * time^2)

In this case, the initial velocity is 27.5 m/s (since the car is traveling at this speed initially), the time is 4.6 seconds (since it takes this time for the car to stop), and the acceleration is the deceleration rate.

However, we do not have the value for acceleration, so we need to find it first.

The formula for deceleration is:

deceleration = (final velocity - initial velocity) / time

In this case, the final velocity is 0 m/s (since the car comes to a stop), the initial velocity is 27.5 m/s, and the time is 4.6 seconds.

Calculating deceleration:
deceleration = (0 - 27.5) / 4.6
deceleration = -27.5 / 4.6
deceleration ≈ -5.98 m/s^2

Now that we have the deceleration value, we can substitute it into the distance formula to calculate the space taken to stop:

distance = (27.5 * 4.6) + (0.5 * -5.98 * 4.6^2)
distance ≈ 126.5 + (-5.98 * 10.6)
distance ≈ 126.5 - 63.49
distance ≈ 63.01 meters

Therefore, it took approximately 63.01 meters for the car to stop.