Question

a car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. if the driver realizes they must stop the car at this time. it takes 0.20 s for the driver to apply the brakes. what must the constant acceleration of the car be after the brakes are applied so that the car would come to rest at the stop sign?

Answer 1

To solve this problem, we can use the kinematic equation:

vf^2 = vi^2 + 2aΔx

Where:
vf = final velocity (0 m/s, since the car comes to rest)
vi = initial velocity (40.0 m/s)
a = constant acceleration
Δx = displacement (200 m)

Plugging in the values, we get:

(0 m/s)^2 = (40.0 m/s)^2 + 2a(200 m)

0 = 1600 m^2/s^2 + 400a

Rearranging the equation, we find:

400a = -1600 m^2/s^2

Dividing both sides by 400:

a = -4.0 m/s^2

Therefore, the constant acceleration of the car after the brakes are applied must be -4.0 m/s^2. Note that the negative sign indicates that the acceleration is opposite to the initial velocity, which is required for the car to come to rest.