A(n) 2153 kg car is coasting along a level

road at 34.4 m/s. A constant braking force
is applied, such that the car is stopped in a
distance of 61.8 m.
What is the magnitude of the braking force?
Answer in units of N

s = 34.4t - 1/2 at^2 = 61.8

v = 34.4 - at = 0

solving for a, a = 9.57 m/s^2

So, now you know that F=ma

To find the magnitude of the braking force, we can use the equations of motion. The key equation we'll use is:

v^2 = u^2 + 2as,

where:
- v is the final velocity (in this case, 0 m/s since the car is stopped),
- u is the initial velocity of the car (34.4 m/s),
- a is the acceleration (which is caused by the braking force), and
- s is the distance over which the car is stopped (61.8 m).

Let's plug in the values we know:

0^2 = (34.4)^2 + 2a(61.8).

Simplifying the equation, we have:

0 = 1183.36 + 123.6a.

Rearranging to isolate the acceleration, we get:

a = -1183.36 / 123.6.

Calculating this, we find:

a ≈ -9.57 m/s^2.

Since the acceleration is negative, it indicates deceleration (opposite to the direction of motion). Now, we can use Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a):

F = ma.

Substituting the given mass of the car (2153 kg) and the acceleration we calculated, we have:

F = 2153 kg × (-9.57 m/s^2).

Calculating this, we find:

F ≈ -20613.21 N.

Since the question asks for the magnitude of the force, we ignore the negative sign:

Magnitude of the braking force ≈ 20613.21 N.

Therefore, the magnitude of the braking force is approximately 20613.21 N.