What average force in N is required to stop an 1,188 kg car in 9 s if it is initially traveling at 89 km/h?

f=ma
f=1,188(9.8)
f=11642.4N

This looks wrong to me. Can someone explain this to me.

Why did you choose 9.8 as the acceleration? It is not falling.

its average velocity during stopping is 89/2 km/hr (change that to m/s)

So it traveled in that time a distance of avgvelocity*9s...

force*distance=initial KE= 1/2 m vi^2
change vi to m/s, then solve for force.

Firstly F=ma but a=(v-u)/t. First convert the 89 Km/h into m/s.

89x1000/3600=24.72m/s

Then using F=m(v-u)/t
F=1188(24.72-0)/9
F=3263.04N

What is the average force need to stop a 1500 N car in 8.0 S if it is traveling at 90 km/hr?

To calculate the average force required to stop a car, we first need to find its initial velocity, final velocity, and the time it takes to stop.

Given:
Mass of the car (m) = 1,188 kg
Initial velocity (u) = 89 km/h
Time taken to stop (t) = 9 s

First, we need to convert the initial velocity from km/h to m/s:
1 km/h = 1000 m/3600 s = 0.2778 m/s

Initial velocity (u) = 89 km/h * 0.2778 m/s = 24.72 m/s

Now, we can calculate the acceleration using the formula:
Acceleration (a) = (final velocity - initial velocity) / time
Since the final velocity is 0 m/s (as the car needs to come to a stop), we have:
Acceleration (a) = (0 - 24.72) / 9
Acceleration (a) = -2.7467 m/s^2

The negative sign indicates that the car is decelerating (slowing down).

Finally, we can find the force using the formula:
Force (F) = mass * acceleration
Force (F) = 1,188 kg * -2.7467 m/s^2
Force (F) = -3,259.05 N

The negative sign indicates that the force is opposing the motion of the car.

So, the average force required to stop the car is approximately -3,259.05 N.