Finding the displacement of a drag racer

Question

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.

JEB · Answer

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

zenaiderrrr · Answer

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) + 1/2 (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=inital velocity, a=acceleration and t=time.
It originates from the formula of s = 1/2(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 X 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 time take.
If I were to substitute v = u + at into s = 1/2(u+v)t, I would get s = ut + 1/2at^2.
That is the explanation. I hope this helps. If you require a diagram to visualise the area under the graph, I will add it in if you ask for it.

Finding the displacement of a drag racer

2 Answers

Add your own answers!

Ask a Question