A 1500 kg car travels along the road at 25 m/s when the driver sees the light turn red and applies the brakes. What average force is needed to stop the car in a) 5 seconds and b) .2 seconds?

What formula should be used?

Force*Time = Initial momentum = 1500*25 kg m/s

Divide 3.75*10^4 by the time interval to get the force, which will be in Newtons.

1500*25=60^2=120

To calculate the average force needed to stop the car, you can use 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.

The formula we will use is:

F = m * a

Where:
F is the force (in Newtons)
m is the mass of the car (in kilograms)
a is the acceleration (in meters per second squared)

Now, let's calculate the average force needed to stop the car in both scenarios:

a) If the car stops in 5 seconds:
In this case, we need to find the acceleration first using the formula:

a = (vf - vi) / t

Where:
vf is the final velocity (which is zero since the car stops)
vi is the initial velocity (given as 25 m/s)
t is the time taken to stop (given as 5 seconds)

Let's calculate the acceleration first:

a = (0 - 25) / 5
a = -5 m/s^2

Now, we can calculate the force using the formula:

F = m * a
F = 1500 kg * (-5 m/s^2)
F = -7500 N (Note: The negative sign indicates that the force is in the opposite direction to the car's motion)

Therefore, the average force needed to stop the car in 5 seconds is 7500 Newtons.

b) If the car stops in 0.2 seconds:
In this case, we again need to find the acceleration first:

a = (vf - vi) / t

Where:
vf is the final velocity (which is zero since the car stops)
vi is the initial velocity (given as 25 m/s)
t is the time taken to stop (given as 0.2 seconds)

Let's calculate the acceleration:

a = (0 - 25) / 0.2
a = -125 m/s^2

Now, we can calculate the force:

F = m * a
F = 1500 kg * (-125 m/s^2)
F = -187,500 N

Therefore, the average force needed to stop the car in 0.2 seconds is -187,500 Newtons.

Note: The negative sign again indicates that the force is in the opposite direction to the car's motion.