A car is travelling along the highway at 90km/h. The driver applies the brakes, the braking action causes a frictional force of 8400 N on the 1050 kg car.

How far does the car travel while it is braking?

Note: answer is 39m but idk how to get there

F = ma

so a = 8400/1050 = -8 m/s^2
90 km/hr = 25 m/s
so
v^2 = 2as
25^2 = 2*8*s
s = 625/16 = 39 m

or, since v = 25-8t, it takes 25/8 seconds to stop, and
s = 25(25/8) - 4(25/8)^2 = 39 m

@oobleck why do we need to multiply v by itself and the acceleration by 2?

BECAUSE

Kinetic energy lost = work done stopping
(1/2) m v^2 = Force * distance = m a * s
so
v^2 = 2 a s

that's what you get if you eliminate t from

v = at
s = 1/2 at^2
note that it also makes the units come out right.

To determine how far the car travels while braking, we can use the equation for linear motion:

distance = (initial velocity^2 - final velocity^2) / (2 * acceleration)

First, let's convert the initial velocity from km/h to m/s. We know that 1 km/h is equal to 1000 m/3600 s. So, the initial velocity is:

initial velocity = 90 km/h * (1000 m/3600 s) ≈ 25 m/s

Next, we need to find the final velocity of the car when it comes to a stop. Since the car is braking, the final velocity will be 0 m/s.

The acceleration can be determined using Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration:

force = mass * acceleration

We have the force applied (8400 N) and the mass of the car (1050 kg). Rearranging the formula, we find:

acceleration = force / mass = 8400 N / 1050 kg ≈ 8 m/s^2

Now we can plug these values into the equation for distance:

distance = (initial velocity^2 - final velocity^2) / (2 * acceleration)
distance = (25 m/s)^2 - (0 m/s)^2 / (2 * 8 m/s^2)
distance = 625 m^2/s^2 / (16 m/s^2)
distance ≈ 39.06 m

Therefore, the car travels approximately 39 meters while it is braking.