A 2500‐kg vehicle traveling at 25 m/s can be stopped by gently applying the breaks for 20 seconds. What is the average force supplied by the breaks to stop the vehicle?

Write "brake", and "braking" when stopping a car.

m=2500 kg
vi=25 m/s
vf=0
t=20 s

vf=vi+aΔt
0=25+a(20)
solve to get
a=-25/20=-5/4=-1.25 m/s²

Use Newton's second law to get force
force, f = ma
=2500 (-1.25) N
=-3125 N

Ignore the negative sign if understood to be braking force.

To find the average force supplied by the brakes to stop the vehicle, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In this case, the vehicle is brought to a stop, so its final velocity is 0 m/s. We can use the equation:

Final velocity (v) = 0 m/s
Initial velocity (u) = 25 m/s
Time taken (t) = 20 s
Mass (m) = 2500 kg

Acceleration (a) is defined as the change in velocity divided by the change in time. In this case, the change in velocity is final velocity (0) minus initial velocity (25):

ΔV = v - u
ΔV = 0 - 25
ΔV = -25 m/s

Now we can calculate the acceleration:

a = ΔV / t
a = -25 / 20
a = -1.25 m/s^2

Since the vehicle is decelerating, the acceleration is negative.

Now, we can calculate the average force (F) using Newton's second law:

F = m * a
F = 2500 * (-1.25)
F = -3125 N

The negative sign indicates that the force is in the opposite direction of the vehicle's motion. Therefore, the average force supplied by the brakes to stop the vehicle is 3125 Newtons in the opposite direction of motion.