Question

In an emergency, a driver brings a car to a full stop in 5 seconds. The car is traveling along ahighway at a rate of 28.8 m/s when braking begins. How far does it travel before stopping?

Answer 1

The car, initially traveling at 28.8 m/s, decelerates uniformly and comes to a full stop in 72 meters after 5 seconds.

Step-by-step explanation:

To determine how far the car travels before stopping when it is initially traveling at 28.8 m/s and comes to a full stop in 5 seconds, we can use the formula for uniform acceleration which is:

s = ut + ½ at²

Where:

• s is the displacement (distance traveled)
• u is the initial velocity (28.8 m/s)
• t is the time (5 s)
• a is the acceleration

Because the car comes to a full stop, the final velocity (v) will be 0 m/s. We can use the equation v = u + at to find the acceleration:

0 = 28.8 m/s + (a)(5 s)

a = -5.76 m/s² (negative sign indicates deceleration)

Now we can calculate the distance using the initial calculated acceleration:

s = (28.8 m/s)(5 s) + ½(-5.76 m/s²)(5 s)²

s = 144 m - 72 m = 72 meters

Therefore, the car travels 72 meters before it comes to a full stop.