Question

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.

Answer 1

The stopping distance of a car initially traveling at 15.0 m/s and decelerating at a rate of 4.00 m/s^2 until it comes to rest is calculated using the kinematic equation. The calculation results in a stopping distance of 28.125 meters.

Step-by-step explanation:

The question involves a car that is initially traveling at 15.0 m/s and decelerates uniformly at a rate of 4.00 m/s2 until it comes to rest. To calculate the stopping distance of the car, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

v2 = u2 + 2as

Where:
v = final velocity (0 m/s since the car comes to rest)
u = initial velocity (15.0 m/s)
a = acceleration (deceleration in this case, so it's -4.00 m/s2)
s = stopping distance (which we need to find)

Substituting the values we have:

0 = (15.0 m/s)2 + 2(-4.00 m/s2)s

Rearranging the equation to solve for s gives us:

s = (15.0 m/s)2 / (2 * 4.00 m/s2)

s = 225.00 m2/s2 / 8.00 m/s2

s = 28.125 meters

Therefore, the stopping distance of the car is 28.125 meters.

Answer 2

28.13 m

Step-by-step 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.