Question

Q2. When the driver applies the brakes of a light truck traveling 40 km/h, it skids 3 m before stopping. How far will the truck skid if it is traveling 80 km/h when the brakes are applied

Answer 1

In physics, the stopping distance is proportional to the square of the speed. So, if the light truck skids 3 meters to stop at 40 km/h, at 80 km/h the truck will skid 12 meters (4 times the original distance) when the brakes are applied, assuming other conditions remain the same.

Step-by-step explanation:

The question relates to the physics of motion, and more specifically, to the concept of deceleration and stopping distances when a force is applied (braking). If a light truck skids 3 meters to stop from a speed of 40 km/h, we can infer that the stopping distance is directly proportional to the square of the speed based on the physics of kinetic energy and braking force.

When the truck is traveling at 80 km/h, which is double the original speed, the stopping distance will be four times the original distance, assuming the same conditions. This is because kinetic energy is proportional to the square of velocity, and when the velocity is doubled, the kinetic energy (and hence the stopping distance under the same deceleration rate) quadruples (22 = 4).

Hence, if the truck skids 3 meters at 40 km/h, it will skid 12 meters if traveling at 80 km/h when the brakes are applied. This example utilizes the basics of kinetic energy and friction to solve practical problems related to driving safety. It serves as a crucial lesson in understanding the impact of speed on the ability to stop a vehicle safely.

Answer 2

12m

Step-by-step explanation:

If the brake force F is constant the brakes have to do work: W = Fx, where x is the skid distance.

The work is necessary to remove the kinetic energy from the truck. It follows:

`E = (1)/(2) mv^(2) = Fx`

m: mass of the truck

v: velocity of the truck

The ratio between the first and the second example would be:

`((1)/(2)mv_1^(2))/((1)/(2)mv_2^(2)) = (Fx_1)/(Fx_2)`

This expression simplifies to:

`(v_1^(2) )/(v_2^(2) ) =(x_1)/(x_2)`

Inserting the speeds will give the ratio for x₁/x₂:

`(x_1)/(x_2) = (40^(2) )/(80^(2) ) = (1)/(4)`

The distance x₂ is 4 times longer then the distance x₁.