Question

A car, of mass \(m\), traveling at a speed \(v_{1}\) can brake to a stop within a distance \(d\). If the car speeds up by a factor of \(2,\) so that \(v_{2}=2 v_{1},\) by what factor is its stopping distance increased, assuming that the braking force \(F\) is approximately independent of the car's speed?

Short answer

Answer: The stopping distance is increased by a factor of 4 when the velocity of the car is doubled.

Step-by-step solution

Write the work-energy principle formula & the formula of work done

The formula for work-energy principle is: W = ΔK where W is the work done, and ΔK is the change in kinetic energy. And the formula of work done is: W = F × d where F represents force and d represents distance.

Express the change in kinetic energy

The change in kinetic energy ΔK can be expressed as follows: ΔK = K_final - K_initial where K_initial is the initial kinetic energy and K_final is the final kinetic energy. Since the car comes to a stop, the final kinetic energy is 0. So, the change in kinetic energy is: ΔK = -K_initial

Write the formula for kinetic energy

The formula for kinetic energy K is: K = (1/2)mv^2 where m is the mass of the car and v is its velocity.

Calculate the initial work done for velocity v1

We can express the initial work done W1 as the product of force F and stopping distance d: W1 = F × d As per the work-energy principle formula, we have W1 = -K_initial: F × d = -(1/2)m(v1)^2

Calculate the initial work done for velocity v2

Since v2 = 2v1, we have: W2 = -K_initial (for v2) F × (n × d) = -(1/2)m(v2)^2 where n is the factor by which stopping distance is increased.

Solve for the factor n

We can now solve for n by dividing equation W2 by equation W1: (F × (n × d)) / (F × d) = (-(1/2)m(v2)^2) / (-(1/2)m(v1)^2) The masses and the force cancel out each other: n = (v2^2) / (v1^2) Since v2 = 2v1, we have: n = ((2v1)^2) / (v1^2)

Calculate the value of n

Finally, we can calculate the value of n: n = (4v1^2) / (v1^2) n = 4 So, the stopping distance is increased by a factor of 4 when the velocity of the car is doubled.

Key concepts

Kinetic Energy

Imagine you're pushing a shopping cart down the aisle. The faster you push it, the harder it is to stop, right? This resistance is due to the cart's kinetic energy, the energy of motion. In physics, kinetic energy is quantified as \( \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

Now, consider a car of mass \( m \) traveling at speed \( v_1 \) which then speeds up to \( v_2 = 2v_1 \). The kinetic energy of the car at speed \( v_1 \) is \( \frac{1}{2}mv_1^2 \), but when the car doubles its speed, the kinetic energy quadruples, since kinetic energy is proportional to the square of the velocity. This exponential growth explains why even a moderate increase in speed can result in a drastic increase in kinetic energy and, in turn, the force required to stop the vehicle.

Stopping Distance

The stopping distance is the total distance a vehicle travels from the moment you realize you need to stop, all the way to a complete halt. It's affected by many factors, such as the road condition, the vehicle's mass, and the initial speed. The exercise mentions a key point: if the car's speed doubles, what happens to the stopping distance?

Using the work-energy principle, which says that the work done (in this case, by the braking force on the car) equals the change in kinetic energy, we can infer that an increased speed results in a higher kinetic energy which requires more work to reduce to zero. This means the car needs a longer distance to stop. As the kinetic energy quadruples when the speed doubles (\( K_2 = 4K_1 \) when \( v_2 = 2v_1 \) ), the stopping distance, assuming constant braking force, will also increase by a factor of four.

Braking Force

Let's talk about the hero of our story: the braking force. This is the force applied by the brakes of the vehicle to bring it to a stop. It does work on the car by converting kinetic energy into other forms of energy, like heat. According to the problem, this force is assumed to be approximately independent of the car's speed.

Using the formula \( W = F \times d \), where \( W \) is the work done by the braking force, \( F \) is the constant braking force, and \( d \) is the stopping distance, we see a direct relationship between the work done to stop the car and the stopping distance. When the work done increases (because of increased kinetic energy from a higher speed), the stopping distance must increase proportionally if the force remains constant. This provides a fundamental insight into the intimate relationship between stopping distance and kinetic energy, governed by the constant braking force.