# Kinetic Energy and Braking Distance [duplicate]

A car traveling at 50km/h will move 44 extra meters from the moment its brakes are applied fully. If the same car travels 100 km/h on the same road, how far will it move from the instant its brakes are fully applied?

I somehow guessed the answer and got it right (176 m) since it was in the Kinetic Energy quiz.

But I'm wondering how are Kinetic Energy and distance are related? And not something like 88m?

I will assume that when the driver "hits the breaks" the force $$F$$ that is acting on the car is constant. In that case, the work $$W$$ that force $$F$$ does is equal to $$W_i = - F d_i \ ,$$ where $$d_i$$ is the distance needed for the car to stop, when the car was initially travelling at a speed $$v_i$$. The negative sign is to recall us that the car is stopping.

The change (final-initial) in the kinetic energy $$T_i$$ is given by $$\Delta T_i = 0 - \frac{1}{2} m v_i^2 \ .$$

Due to conservation of energy, we know that the kinetic energy loss is equal to the work of friction, i.e. $$W_i = \Delta T_i$$

One way to solve the problem is to compute the quotient between the two cases in the question (a): $$\frac{W_2}{W_1} = \frac{\Delta T_2}{\Delta T_2}$$

Replacing with the respective expressions we get $$\frac{-F d_2}{-F d_1} = \frac{-\frac{1}{2} m v_2^2}{-\frac{1}{2} m v_1^2}$$

Simplifying the quotients gives: $$\frac{d_2}{d_1} = \frac{v_2^2}{v_1^2}$$

Using this last expression will allow us to get the answer to the question.

(a)

The quotient makes sense because the variation of the kinetic energy is equal to the work done by friction both in case 1 and in case 2. This means that, in practice, we are dividing both sides of the equation by the same number. For instance, if $$x=y$$, then $$x/2 = y/2$$; moreover, if $$x=y$$ and $$a=b \neq 0$$, then we have that $$x/a = y/a \iff x/a = y/b$$.

Another, perhaps more complicated way to see that the quotient makes sense is the following: $$W_i = \Delta T_i \iff \frac{W_i}{\Delta T_i} = 1 \ .$$

Hence, $$\frac{W_2}{\Delta T_2} = 1 \text{ and } \frac{W_1}{\Delta T_1} = 1 \ .$$

Therefore, $$\frac{W_2}{\Delta T_2} = \frac{W_1}{\Delta T_1} \iff \frac{W_2}{W_1} = \frac{\Delta T_2}{\Delta T_1} \ .$$

• Finally makes sense. But I’m not sure why the quotient are equal – 003264 _ Jan 11 at 2:30
• In this example we have two systems of equations. One for each initial speed. From this we can manipulate the equations by mathematical rules to obtain our result. – Alexander Issa Jan 11 at 11:21
• @003264_ I've added a footnote to my post, trying to make the explanation clearer. – Ertxiem - reinstate Monica Jan 11 at 20:58

This is an example of work-energy theorem. It states, that the change in kinetic energy is equal to work done on an object. $$W= \Delta T\\\text{where: T}=\text{kinetic energy}$$ The quantity you are given are initial and final kinetic energy and distance travelled by decelerating car. Since it has a mass and it's velocity changes, force must act on it by Newton's second Law.

This information should allow you to get answer without involving gambling.

• But should not friction force be non-conservative? – Leo Liu Jan 11 at 1:59
• I was confused on how to compute the final velocity, then I see a solution setting it equal to zero and I’m like “how did I not realise that” I’ll give you an upvote – 003264 _ Jan 11 at 2:33
• You are right. In realistic case, we should take into account that the breaking force is not constant and mechanical energy gets transformed into heat. Also question arises if vehicle wheels are rolling, their moment of inertia etc. However this is a simplified model. As for non conservative force, if only mechanical energy is considered, the work done by this force is specifically equal to change in T. I have posted answer regarding this theorem here: physics.stackexchange.com/questions/469651/… – Alexander Issa Jan 11 at 11:29

Increasing speed squares the kinetic energy. So doubling the car's speed would not be 2 times the kinetic energy, it would be 2 squared times the kinetic energy. So 4 times the kinetic energy would require 4 times the braking distance.

• Doesn’t exactly show the relationship between kinetic energy and distance – 003264 _ Jan 11 at 2:37