The driver of a 1.00X103 kg car traveling on the interstate at 35.0 m/s slams on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the brakes are applied, a constant friction force of 8.00X103 N acts on the car. Ignore air resistance.

(a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collision occur?

the car has kinetic energy ... 1/2 m v^2

to stop the car; the work done by the brakes (f * d) must equal the car's kinetic energy

(a) 1/2 * 1.00E3 * 35.0^2 = 8.00E3 * d

(b) find the work done in 30.0 m and subtract it from the initial K.E.
... the remaining energy will give the collision speed

(a) Well, we certainly don't want the driver to have an unexpected collision with the other vehicle. That would definitely put a dent in their driving record! To avoid any unpleasant surprises, let's calculate the minimum distance required for the car to come to a stop before reaching the other vehicle.

First, let's find the deceleration of the car. We can use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration (F = ma). Rearranging the equation, we find that acceleration (a) is equal to the net force (F) divided by the mass (m).

Given that the mass of the car is 1.00x10^3 kg and the friction force acting on it is 8.00x10^3 N, the acceleration can be calculated as follows:

a = F/m
a = 8.00x10^3 N / 1.00x10^3 kg
a = 8.00 m/s^2

Now, we can use the kinematic equation v^2 = u^2 + 2as, where v is the final velocity of the car (0 m/s, since it needs to come to a stop), u is the initial velocity (35.0 m/s), a is the acceleration (-8.00 m/s^2, since we want the car to slow down), and s is the distance we're trying to find.

Substituting the known values into the equation:

0^2 = 35.0^2 + 2(-8.00) * s

Simplifying the equation:

0 = 1225.0 - 16s

Rearranging the equation and isolating s:

16s = 1225.0
s = 1225.0 / 16
s ≈ 76.6 m

So, the minimum distance required for the car to avoid a collision is approximately 76.6 meters.

(b) If the distance between the vehicles is initially only 30.0 m, and we're assuming the driver didn't apply the brakes early enough, let's calculate the speed at which the collision would occur.

Using the same kinematic equation v^2 = u^2 + 2as, we can find the final velocity (v) at the moment of collision. Now, u is the initial velocity of the car (35.0 m/s), a is the acceleration (-8.00 m/s^2), and s is the initial distance between the vehicles (30.0 m). We can solve for v as follows:

v^2 = 35.0^2 + 2(-8.00) * 30.0

Simplifying the equation:

v^2 = 1225.0 - 480.0
v^2 = 745.0

Taking the square root of both sides:

v ≈ √745.0
v ≈ 27.3 m/s

Therefore, if the distance between the vehicles is initially only 30.0 meters, the collision would occur at a speed of approximately 27.3 m/s. It's always best to maintain a safe distance on the road and apply the brakes early enough to avoid any unnecessary fender benders. Safety first, after all!

To solve this problem, we can use the equations of motion to determine the minimum distance required to avoid a collision (part a) and the speed at which the collision would occur (part b).

Part a:
We need to find the minimum distance required to stop the car before hitting the second vehicle. We can use the equation of motion that relates distance, initial velocity, final velocity, acceleration, and time:

vf^2 = vi^2 + 2ad

where:
- vf: final velocity is 0 m/s since the car needs to come to a stop
- vi: initial velocity is 35.0 m/s
- a: acceleration is given by the friction force divided by the mass of the car, a = F_f / m
- d: distance is what we need to find

First, we calculate the acceleration:
a = F_f / m
a = 8.00x10^3 N / (1.00x10^3 kg)
a = 8.00 m/s^2

Now we can solve for the minimum distance:
0^2 = (35.0 m/s)^2 + 2(8.00 m/s^2)d
0 = 1225 m^2/s^2 + 16.00 m/s^2 d
-1225 m^2/s^2 = 16.00 m/s^2 d
d = -1225 m^2/s^2 / (16.00 m/s^2)
d = -76.56 m/s^2 (which is the minimum distance needed to stop)

However, the distance cannot be negative, so we take the positive value:
d = 76.56 m/s^2

Therefore, the minimum distance the brakes should be applied to avoid a collision is 76.56 meters.

Part b:
If the distance between the vehicles is initially only 30.0 meters, then a collision would occur if the time taken to cover this distance is less than the time required to stop the car.

To determine the speed at which the collision would occur, let's calculate the time needed to cover the distance of 30.0 meters:

d = vi*t + 0.5*a*t^2

where:
- d: distance is 30.0 m
- vi: initial velocity is what we need to find
- a: acceleration is 8.00 m/s^2
- t: time is what we need to find

Rearranging the equation, we get:

0.5*a*t^2 + vi*t - d = 0

Using the quadratic formula to solve for t:

t = (-b ± √(b^2 - 4ac)) / (2a)

where:
- a = 0.5*a = 4.00 m/s^2
- b = vi = initial velocity
- c = -d = -30.0 m

Plugging the values into the quadratic formula:

t = (-vi ± √(vi^2 - 4(4.00)(-30.0))) / (2(4.00))

We ignore the negative value of t since time cannot be negative. Therefore:

t = (-vi + √(vi^2 + 480)) / 8.00

To avoid a collision, the time taken to cover the distance should be greater than or equal to the time required to stop the car. Therefore:

(-vi + √(vi^2 + 480)) / 8.00 ≥ 35.0 / 8.00

Solving this inequality will give us the range of possible velocities at which the collision would occur.

I'm sorry, but I am unable to solve inequalities and provide a specific answer for part b of your question.

To find the minimum distance required for the car to come to a stop and avoid collision, we can use the equations of motion.

(a) First, let's find the deceleration of the car using the given information. The net force acting on the car is the friction force, which is given as 8.00X10^3 N. According to Newton's second law of motion, the net force is equal to the mass of the car multiplied by its acceleration:

Fnet = m * a

Rearranging the equation, we get:

a = Fnet / m

Substituting the given values, we have:

a = (8.00X10^3 N) / (1.00X10^3 kg)

a = 8.00 m/s^2

Now, let's use the equations of motion to find the minimum stopping distance.

The equation that relates the initial velocity (Vi), final velocity (Vf), acceleration (a), and displacement (d) is:

Vf^2 = Vi^2 + 2ad

In this case, the initial velocity (Vi) is 35.0 m/s, the final velocity (Vf) is 0 m/s (since the car comes to a stop), and the acceleration (a) is -8.00 m/s^2 (negative since it is in the opposite direction to the car's initial motion).

Plugging in the values, we have:

0^2 = (35.0 m/s)^2 + 2(-8.00 m/s^2) * d

Rearranging the equation, we get:

30.0 m/s^2 * d = (35.0 m/s)^2

d = (35.0 m/s)^2 / (2 * 8.00 m/s^2)

d = 61.6 meters

Therefore, the minimum distance required for the car to avoid the collision is 61.6 meters.

(b) If the initial distance between the vehicles is only 30.0 meters, and the car requires a minimum stopping distance of 61.6 meters, we can determine whether a collision will occur.

If the stopping distance is greater than the initial distance, a collision will occur. On the other hand, if the stopping distance is less than the initial distance, there will be no collision.

In this case, the stopping distance (61.6 meters) is greater than the initial distance (30.0 meters), so a collision will occur.

To calculate the speed at which the collision will occur, we can again use the equations of motion. Since the car comes to rest after the collision, the final velocity (Vf) will be 0 m/s. We need to find the initial velocity (Vi).

The equation that relates the initial velocity (Vi), final velocity (Vf), acceleration (a), and displacement (d) is:

Vf^2 = Vi^2 + 2ad

In this case, the acceleration (a) is the same as before (-8.00 m/s^2), the displacement (d) is 30.0 meters, and the final velocity (Vf) is 0 m/s.

Plugging in the values, we have:

0^2 = Vi^2 + 2(-8.00 m/s^2) * 30.0 m

0 = Vi^2 - 480 m^2/s^2

Vi^2 = 480 m^2/s^2

Vi = √480 m/s

Vi ≈ 21.9 m/s

Therefore, if the initial distance is only 30.0 meters, the collision will occur at a speed of approximately 21.9 m/s.