# This question is similar to some on driver's-license exams: A car moving at 50 km/h skids 15 m with locked brakes. Show that with locked brakes at 150 km/h the car will skid 135 m.

## d = d1(V2/V1)^2 = 15(150/50)^2 = 135 m.

## To solve this problem, let's start by calculating the braking distance of the car moving at 50 km/h. We can use the following formula:

Braking Distance = ((Initial Velocity)^2) / (2 * Deceleration)

Given that the car skids 15 m and the initial velocity is 50 km/h, we need to convert the velocity to m/s:

50 km/h = 50 * (1000 m / 3600 s) = 13.89 m/s (approximately)

Next, we need to calculate the deceleration. It is given by the formula:

Deceleration = (Final Velocity^2 - Initial Velocity^2) / (2 * Braking Distance)

Given that the final velocity is 0 (since the car comes to a stop), the deceleration becomes:

Deceleration = (0 - (13.89 m/s)^2) / (2 * 15 m)

Deceleration = (-13.89 m/s)^2 / (2 * 15 m)

Deceleration = 96.83 m^2/s^2 / 30 m

Deceleration = 3.23 m/s^2

Now that we have the deceleration, we can calculate the braking distance for the car moving at 150 km/h. Using the formula mentioned earlier:

Braking Distance = ((Final Velocity)^2) / (2 * Deceleration)

First, we need to convert the velocity to m/s:

150 km/h = 150 * (1000 m / 3600 s) = 41.67 m/s (approximately)

Then, calculating the braking distance:

Braking Distance = (41.67 m/s)^2 / (2 * 3.23 m/s^2)

Braking Distance = 1734.72 m^2/s^2 / 6.46 m/s^2

Braking Distance ≈ 268.34 m

Hence, with locked brakes at 150 km/h, the car would skid approximately 268.34 m, not 135 m.

## To find the relationship between the speeds and distances traveled with locked brakes, we can use the concept of kinetic energy. Kinetic energy is given by the equation:

KE = (1/2)mv^2

Where KE is the kinetic energy, m is the mass of the object, and v is the velocity.

Since the mass of the car remains constant, we can ignore it for this problem and focus only on the change in kinetic energy due to the change in velocity.

When the car's speed changes from 50 km/h to 150 km/h, the ratio of the kinetic energies can be expressed as:

(KE2 / KE1) = (v2^2 / v1^2)

Substituting the given values:

(KE2 / KE1) = (150^2 / 50^2)

Simplifying the equation:

(KE2 / KE1) = (22500 / 2500)

(KE2 / KE1) = 9

This means the kinetic energy of the car with locked brakes at 150 km/h is nine times greater than when it is traveling at 50 km/h.

Now, the change in kinetic energy is directly proportional to the distance traveled during a skid. We can represent this relationship as:

Δd = k * ΔKE

where Δd is the change in distance, k is a proportionality constant, and ΔKE is the change in kinetic energy.

Since we know that the car skids 15 m at 50 km/h, we can set up the equation:

15 = k * ΔKE1

And for the car traveling at 150 km/h:

Δd = k * ΔKE2

Substituting the values:

Δd = k * 9 * ΔKE1

Now we can solve for Δd when the car is traveling at 150 km/h:

Δd = 9 * 15

Δd = 135 m

Therefore, with locked brakes at 150 km/h, the car will skid 135 m.