Answer:
a) 4116.67 N
b) negative (assuming car is traveling in positive direction)
c) 19.125 m
Step-by-step explanation:
Part A)
We can find the average force exerted on the car by using Newton's 2nd Law: F = ma.
We have the mass (950 kg) but we do not have the acceleration. Let's solve for this using kinematics.
We have the time during braking, the initial and final velocity, and we need to find the acceleration. To do this, we can use the following equation: v = v₀ + at.
- v (final velocity) = 9.50 m/s
- v₀ (initial velocity) = 16.0 m/s
- t (time) = 1.50 s
- a = ?
Substitute these values into the equation.
- v = v₀ + at
- 9.50 m/s = 16.0 m/s + a (1.50 s)
- 9.50 = 16.0 + 1.50a
- -6.50 = 1.50a
- a = -4.333333... m/s²
The car's acceleration during the braking period is -4.33 m/s², assuming the car is traveling in the positive direction.
Now, we can use this to plug into Newton's 2nd Law to find the force exerted on the car.
- F = ma
- F = (950 kg) (-4.33...)
- F = -4116.666667 N
The magnitude of the average force exerted on the car while braking is the absolute value, so it is 4116.67 N.
Part B)
The direction of the average force is denoted in the negative sign. It is opposite the direction of the car's motion. Assuming that the car is traveling in the positive direction, the direction of the average force exerted on the car is in the negative direction since this force is slowing the car down while braking.
Part C)
We can find the distance traveled during the braking period by using another kinematic equation.
We have the initial and final velocity, and the time traveled, so we can use this equation: Δx = v_avg · t.
- Note that v_avg is the same thing as (v + v₀) / 2.
Substitute the known values into the equation.
- Δx = [(9.50 + 16.0)/2] · 1.50
- Δx = 12.75 · 1.50
- Δx = 19.125 m
The car traveled 19.125 m while braking.