# A 1000 kg car traveling is at 40.0 m/ s. To avoid hitting a second car the driver slams on his brakes.

The brakes provide a constant friction force of 8000 N. ( a) Find the minimum distance that the

brakes should be applied in order to avoid a collision with the vehicle in front? ( b) What would be

the speed of collision if the distance between the two vehicles is initially 45.0 m.

## a. F = M*a.

a = F/M = -8000/1000 = -8m/s^2.

V^2 = Vo^2 + 2a*d.

0 = 40^2 - 16d

d = 100m. to stop.

b. V^2 = Vo^2 + 2a*d.

V^2 = 40^2 - 16*45 =

V = ?.

## Helpful

## F = m a

-8000 = 1000 a

a = -8 m/s^2

v = 40 - 8 t

we want v = 0

8 t = 40

t = 5 seconds to stop (yikes!)

d = 40*5 - (1/2)(8)(25)

= 200 - 100

= 100 meters

for part b

how long to for 45 m?

45 = 40 t -4 t^2

solve quadratic for t

then

v = 40 - 8 t

## a) Well, if the driver wants to avoid a collision, they better apply those brakes pronto! To find the minimum distance needed to avoid a collision, we need to consider the forces at play.

The net force acting on the car in the direction of motion when the brakes are applied is given by Newton's second law as:

Net force = mass × acceleration

Since the car wants to decelerate (i.e., change its speed in the opposite direction of motion), the acceleration is negative:

Net force = mass × (-acceleration)

The force due to friction provided by the brakes is acting in the opposite direction of motion, so it's also negative:

Net force = (-1) × friction force

Setting these two equal, we have:

mass × (-acceleration) = (-1) × friction force

Rearranging the equation to solve for acceleration, we get:

acceleration = friction force / mass

Plugging in the given values, we have:

acceleration = 8000 N / 1000 kg

acceleration = 8 m/s²

The negative sign indicates deceleration. Now, let's use the kinematic equation to find the minimum distance (d) needed to stop:

v² = u² + 2as

Where:

v = final velocity (0 m/s when stopped)

u = initial velocity (40.0 m/s)

a = acceleration (-8 m/s²)

s = distance

Plugging in the values:

0² = 40.0² + 2(-8)s

0 = 1600 - 16s

16s = 1600

s = 100 m

Therefore, the minimum distance needed to avoid a collision is 100 meters.

b) Now let's calculate the speed of the collision if the distance between the two vehicles is initially 45.0 m. In this case, the car doesn't have enough distance to stop, so a collision is inevitable. But hey, why so serious? Let's calculate it anyway!

Using the same kinematic equation as before, but this time solving for final velocity, we have:

v² = u² + 2as

Where:

v = final velocity (what we want to find)

u = initial velocity (40.0 m/s)

a = acceleration (-8 m/s²)

s = distance (45.0 m)

Plugging in the values:

v² = 40.0² + 2(-8)(45.0)

v² = 1600 - 720

v² = 880

v ≈ ± 29.66 m/s

Well, it seems like we're getting a bit of a complex answer. We should take the positive value for the final velocity, as a negative velocity would mean the car is moving backward (which doesn't make sense in this context).

Therefore, the speed of the collision would be approximately 29.66 m/s. Just remember to avoid colliding with clowns whenever possible!

## To solve this problem, we can use the principles of physics, specifically the equations of motion and the concept of work and energy.

(a) To find the minimum distance that the brakes should be applied in order to avoid a collision, we need to determine the distance the car would travel while decelerating from its initial speed to a stop.

The formula we will use is:

v² = u² + 2as

Where:

- v is the final velocity (0 m/s since the car stops)

- u is the initial velocity (40.0 m/s)

- a is the acceleration (which we need to find)

- s is the distance traveled

Rearranging the equation, we have:

a = (v² - u²) / (2s)

Substituting the given values:

a = (0² - 40.0²) / (2 * s)

Since the acceleration is given by the frictional force divided by the mass of the car:

a = F / m

We can substitute this into the equation:

F / m = (0² - 40.0²) / (2 * s)

Simplifying further:

8000 N / 1000 kg = (-40.0 m/s)² / (2 * s)

Now we can solve for s:

s = ((-40.0 m/s)² / (2 * 8000 N / 1000 kg))

Calculating this, we get:

s ≈ 80.0 m

Therefore, the minimum distance that the brakes should be applied to avoid a collision is approximately 80.0 meters.

(b) To find the speed of the collision if the distance between the two cars is initially 45.0 meters, we can use the principle of conservation of energy.

The work done by the braking force is equal to the car's initial kinetic energy. The work done on an object can be calculated as the force multiplied by the distance:

Work = Force x Distance

The work done by the braking force is:

Work = 8000 N x 45.0 m

Since work is equal to the change in kinetic energy:

Work = ΔKE = (1/2)mv² - (1/2)mu²

Rearranging the equation and substituting the given values:

(1/2)(1000 kg)v² - (1/2)(1000 kg)(40.0 m/s)² = 8000 N x 45.0 m

Simplifying further:

(1/2)(1000 kg)v² = (1/2)(1000 kg)(40.0 m/s)² + 8000 N x 45.0 m

Now we can solve for v:

v² = [(1/2)(40.0 m/s)² + (8000 N x 45.0 m)] / (1000 kg)

Calculating this, we get:

v ≈ 30.0 m/s

Therefore, the speed of the collision would be approximately 30.0 m/s.