Physics Related... Help?

tannawirsch:

Okay, so I am having a huge struggle with some Physics questions from Uniform Accelerated Motion.

There are three equations that I can use:

1) V₂ = V₁ + a∆t
2) ∆d = V₁∆t + 1/2 a∆t^2
3) V₂^2 = V₁^2 + 2 a∆d

I am having troubles solving a question using equations (2). Say for example:

A sports car approaches a highway on-ramp at a velocity of 20.0m/s [E]. If the car accelerates ata rate of 3/2m/s^2 [E] for 5.0s, what is the displacement of the car?

The answer is 100m [E] + 40m [E]
∆d = 1.4 x 10^2m [E]

I know that you would add in your variables: ∆d = V₁(20.0m/s) ∆t(5.0s) + 1/2 a(3.2m/s[E]) ∆t(5.0)^2
I know how to get the 100m: (20.0)*(5.0)=100

But how do you get the 40?
Or even how do you get ∆d = 1.4 x 10^2 ?

You were on the right track by formulating ∆d = V₁(20.0m/s) ∆t(5.0s) + 1/2 a(3.2m/s[E]) ∆t(5.0) ²

Yet, what you haven't done is actually filling in the numbers.

This gives ∆d = 20.0*5.0+ (1/2)*3.2 *5.0 ² = 100 + 1.6*5 ² = 100 + 1.6*25 =
100+ 40 =
140 =
1.4* 100 =
1.4 * 10 ²

So, the 20*5 = 100 gives the distance the car travels due to its beginning speed,
and the (1/2)*3.2 *5.0 ² gives the extra distance the car travels due to the extra speed it gains from accelerating.

You calculated the first part by your self but missed the accelerating part.

also 140 = 1.4 * 10 ² which is in essence the same.

The difference is that 140 describes the distance with 3 significant figures while 1.4 * 10 ² describes it with 2 significant figures

If you want to learn more about that look at;

http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/index.html


Reinout

Awesome! Thanks I had the worst time trying to figure that out, LOL.