A car is originally traveling at 15.0 m/s on a straight horizontal road. The driver applies the brakes, causing a car to decelerate uniformly at 4.00 m^2 until it comes to rest. Calculate the car's stopping distance.

28.13 m

Explanation:

Parameters given:

Initial speed of the car, u = 15 m/s

Acceleration of the car, a = -4 m/s ²

(The acceleration is negative because it is a deceleration)

We can find the distance traveled by the car by applying of the Newton's equations of motion:

v² = u² + 2 * a * s

Where v = final velocity of the car (0 m/s since the car comes to rest)

s = distance moved by the car

Therefore:

0² = 15² + (2 * -4 * s)

0 = 225 - 8s

=> s = 225/8

s = 28.13 m

The distance moved by the car before it comes to a stop is 28.13 m.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

the ima (higher than 1m) will be greater than the ima with the height of 1m.