# A car weights 1.30 X 10^4 N is initially moving at a speed of 40km/h when the brakes are applied and the car is brought to a stop in 15m. Assuming that the force that srops the car is constant. find a)magnitude of the force b)time required for the change in speed.

If the inital speed is doubled and the car experiances the same force during the braking by what factors are c) the stopping distance and d) the stopping time multiplied?

I have a test with this kind of problem on it soon and I still don't understand it. Please help me.

Use the impulse equation:

Force*time= mass*changeinvelocity

## To solve this problem, we will use the impulse equation:

Impulse = Force * time = mass * change in velocity

a) Magnitude of the force:
Given:
Initial speed (u) = 40 km/h = 40,000 m/3600 s = 11.11 m/s (converted from km/h to m/s)
Final speed (v) = 0 m/s (the car comes to a stop)
Change in velocity (Δv) = v - u = 0 - 11.11 = -11.11 m/s (negative because the car is slowing down)

The impulse equation can be rearranged to solve for force:
Force = Impulse / time

Since the car is brought to a stop, the impulse is equal to the change in momentum, which is the mass times the change in velocity:
Impulse = mass * Δv

Substituting the given values:
Force = (mass * Δv) / time

Given that the weight of the car is 1.30 x 10^4 N, we can convert this to mass using Newton's second law (weight = mass * gravity):
mass = weight / gravity

Using the value of gravity as 9.8 m/s^2 (standard value):
mass = (1.30 x 10^4 N) / (9.8 m/s^2) = 1326.53 kg (approx.)

Now we have all the values to substitute into the force equation:
Force = (1326.53 kg * -11.11 m/s) / 15 m
Force = -972.02 N

The magnitude of the force is 972.02 N (since force is a vector and magnitude represents its size without considering direction).

b) Time required for the change in speed:
Given: change in velocity (Δv) = -11.11 m/s (negative because the car is slowing down)

We can rearrange the impulse equation to solve for time:
time = (mass * Δv) / force
time = (1326.53 kg * -11.11 m/s) / (-972.02 N)
time ≈ 15.21 s

The time required for the change in speed is approximately 15.21 seconds.

c) Stopping distance when initial speed is doubled:
If the initial speed is doubled, the new initial speed (u') would be 2 * 11.11 m/s = 22.22 m/s.

Using the same impulse equation, we can calculate the new stopping distance (d'):

Force * time = mass * change in velocity
Force * d' = mass * (0 - 22.22)

Since the force applied remains the same, we can write:
-972.02 N * d' = 1326.53 kg * -22.22 m/s

Simplifying the equation:
d' = (1326.53 kg * 22.22 m/s) / 972.02 N
d' ≈ 30.31 m

The stopping distance is approximately 30.31 meters when the initial speed is doubled.

d) Stopping time when initial speed is doubled:
To calculate the stopping time (t') when the initial speed is doubled, we can use the same equation:

Force * time = mass * change in velocity
-972.02 N * t' = 1326.53 kg * (0 - 22.22)

Simplifying the equation:
t' = (1326.53 kg * 22.22 m/s) / 972.02 N
t' ≈ 30.31 s

The stopping time is approximately 30.31 seconds when the initial speed is doubled.

To recap:
a) The magnitude of the force is 972.02 N.
b) The time required for the change in speed is approximately 15.21 seconds.
c) The stopping distance is approximately 30.31 meters when the initial speed is doubled.
d) The stopping time is approximately 30.31 seconds when the initial speed is doubled.