# Finding the displacement of a drag racer

Sorry for asking such an obtuse question but I have no one to resort to about the following explanation for the answer below. I'm self-teaching myself physics so please don't be so hard on me and please give a simple answer as if I am 12. The part I'm having trouble with is why I have to increase the displacement by a factor of 5^2, or 25. Thanks for you help! Here is the prompt:

A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be?

STRATEGIZE We assume that the acceleration is constant. Because the initial position and velocity are zero, the displacement will then scale as the square of the time; we can then use ratio reasoning to solve the problem.

PREPARE After 1.0 s, the car has traveled 6.0 m; after another 4.0 s, a total of 5.0 s will have elapsed.

SOLVE The initial elapsed time was 1.0 s, so the elapsed time increases by a factor of 5. The displacement thus increases by a factor of 5^2 , or 25. The total displacement is ∆x = 2516.0 m2 = 150 m

ASSESS This is a big distance in a short time, but drag racing is a fast sport, so our answer makes sense.

Looks ok. It's easier, but equivalent, to draw $$v(t) = 0 + at$$, which is a triangle.

The slope:

$$\frac{dv}{dt} = a$$

is the (constant) acceleration, while the integral:

$$\int_0^t{v(t')dt'} = \frac 1 2(base)\times(height)=\frac 1 2 t (at)=\frac 1 2 at^2$$

is the displacement. Then, similar triangles give you:

$$\frac{d_2}{d_1}=\big(\frac{t_2}{t_1}\big)^2=25=\frac{\Delta x}{6\,m}$$

• Sorry, those comments weren't actually mine, I took them out of the textbook. I still don't see why it has to be 5^2 regardless of how much time I spent trying to make sense of basic physics. I appreciate your help, anyway. Hope you have a good evening! Commented Mar 28, 2021 at 5:25
• @Nathan I got that, about the square. That's why I suggested you draw a triangle (which is 1/2 a rectangle, and a rectangle can be rescaled into square).
– JEB
Commented Mar 28, 2021 at 14:30

So initial velocity = $$0m/s$$.

Total time taken = $$1s + 4s = 5s.$$

Acceleration, assuming it is constant = $$6m/s^2.$$

How far will the car be from the starting line = $$displacement = ut + 1/2 at^2 = (0m/s)(5s) + \frac12 (6m/s^2)(5s)^2 = 75m$$

Your question I suppose is why must a square the time? For this, we need to understand the derivation of this formula of $$s = ut + 1/2at^2.$$ where $$s=displacement,\ u=initial\ velocity,\ a=acceleration\ and\ t=time.$$

It originates from the formula of $$s = \frac12(u+v)t.$$ This is the formula of a trapezium. It represents the area under the graph of a velocity-time graph. If I were to find the are under the graph of a velocity-time graph, Im actually finding the displacement because $$velocity\ *\ time = displacement.$$

Now you also need to know the formula $$v = u + at.$$ The final velocity, $$v$$, is equals to the initial velocity plus the acceleration and the time taken.

If I were to substitute $$v = u + at$$ into $$s = \frac12(u+v)t$$, I would get $$s = ut + \frac12at^2.$$

That is the explanation. I hope this helps. If you require a diagram to visualize the area under the graph, I will add it in if you ask for it.