Homework Help: This Must B Da Most Easiest Phy Prob.

1. Jul 12, 2004

scisyhp

I feel so dump. I've never taken physic in high school, and now I'm struggling on my second problem. And it's only the beginning of the term . Can someone pls help.

If a person on a bike is at rest and pushes himself forward with a constant acceleration of 4.0m/s^2. what is his speed after 6.0?

that would be:V=12.0m/s

but this is the one that really getting to me.
For an additional 6.0s, at the same constant speed, the biker forgot his shoes and presses the brakes that bring him to a stop at 5.0s. What is his acceleration? Is it positive

*I've been working on this problem for more than 10hours, and I'm just frustrated at not knowing.I've tried all the formulas. I understand the constant acceleration but when it gets to the part where he presses the brake its really tricky cause there is only time. :surprise: :grumpy: No mention of distance, velocity, and etc.
Can anyone help..pls ?

2. Jul 12, 2004

AKG

You have time, as you know. You also have initial velocity, as it is the velocity you calculated in the part above (assuming you did it right). You have final velocity, zero, because the guy brakes and comes to a stop.

3. Jul 12, 2004

scisyhp

the first question I put down the wrong answer it should be-
V=24.0m/s

Second problem
t=10s
a=4.0m/s^2
V=40m/s
but the brake is still confusing me:
which formula do I use, to solve for it? and how do I arrive to the correct answer?

4. Jul 12, 2004

AKG

I have no idea what you're doing.

First problem:

$a = 4.0 m/s^2$
$v_i = 0 m/s$
$t = 6.0 s$
$v_f = at + v_i = 24 m/s$

Second Problem:

$v_i = 24 m/s$
$t = 5.0 s$
$v_f = 0 m/s$
$a = (v_f - v_i)/t = -24/5.0 = -4.8 m/s^2$

5. Jul 12, 2004

Gza

Solving physics problems isn't about randomly plugging numbers into formulas. If you find yourself spending more than *15-20 minutes* on a problem without writing anything or going anywhere, than open up your book and read the chapter carefully. You obviously didn't understand the concepts of velocity and acceleration well enough to solve the problem, so don't stress yourself out if you can't do it.

6. Jul 12, 2004

scisyhp

Gza- well the thing is I have used my book as a reference, and looking at all the examples inside the book. The thing is I'm just confused with the additional seconds and how I kept arriving at higher numbers than what I thought it might be.

AKG- Gosh your so nice to take your time up to solve my problem. What I'm just a little confused about is-
So I don't need to add the additional 6.0s?but why not? wouldn't that make the problem wrong?
Because he went an extra sec before pressing his brakes?

7. Jul 12, 2004

JohnDubYa

The question is badly phrased. Either they want the average acceleration for the entire six seconds (which is simply the change in speed divided by six seconds) or the instantaneous acceleration for the first five seconds (which is the change in speed divided by five seconds).