A motor car of mass 30kg,moving with a velocity of 10m/s. by the application of brakes,it was brought to rest at a distance of 25meters. Find the average force.

F=ma=mv²/2s

I have no idea

To find the average force exerted by the brakes, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

First, we need to calculate the acceleration of the car using the given information. The car starts with an initial velocity of 10 m/s and comes to rest (final velocity is 0 m/s). Therefore, the change in velocity is 10 m/s (initial velocity - final velocity).

We can use the following equation to relate acceleration, change in velocity, and distance:
v^2 = u^2 + 2as

Where:
v = final velocity (0 m/s)
u = initial velocity (10 m/s)
a = acceleration
s = distance (25 meters)

Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)

Substituting the given values, we have:
a = (0^2 - 10^2) / (2 * 25)
a = -100 / 50
a = -2 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the velocity. In this case, since the car is decelerating (braking), the acceleration is negative.

Now that we have the acceleration, we can calculate the average force using Newton's second law:
F = m * a

Substituting the given mass of the car (30 kg) and the calculated acceleration (-2 m/s^2), we have:
F = 30 kg * (-2 m/s^2)
F = -60 N

The negative sign indicates that the force is acting in the opposite direction of motion, which makes sense as the brakes are applied to slow down the car.

Therefore, the average force exerted by the brakes is 60 Newtons in the opposite direction of the car's motion.