Equations for Accelerated Motion 4

1. Oct 20, 2013

Medgirl314

1. The problem statement, all variables and given/known data
A motorcycle traveling at 30 m/s comes to a stop over a distance of 150 m. What was the motorcycle's acceleration?

2. Relevant equations
v^2=vinitial+2aΔx

3. The attempt at a solution
I tried plugging in what I know.

0=900+2a*150 m

Then I divided 150 m and 900 by 150:

0=6+2a

This yields an answer of -4. It doesn't seem right, but I can't place why. I think I may have needed to divide the v^2 by 150, but that's 0/150, which wouldn't make sense. Thanks in advance for a reply!

2. Oct 20, 2013

NihalSh

does it?....check your final equation, you have done the rest of it correctly....it should be -ve because of deceleration (retardation), it just shows acceleration acts in direction opposite to its initial velocity

$0=6+2a$
$a=-3$

Edit: I think it was a typo, but this is what the equation is v^2=(vinitial)^2+2aΔx.....your calculations are correct except the very end of it.!!!

Last edited: Oct 20, 2013
3. Oct 20, 2013

Medgirl314

Oh! Okay. I made a mistake in the post, I actually thought it was -2 m/s . I see what you did there, I added in my head instead of dividing. The main reason I thought it was wrong didn't have a reason. It just seemed like it should be different because I couldn't place how an acceleration of -2 or -3 would be enough to slow the car, but now that I look again, it makes sense. Thank you for a polite and helpful reply!

4. Oct 20, 2013

Medgirl314

The car?? I meant the motorcycle. Sorry. Blame it on lack of coffee syndrome.

5. Oct 20, 2013

NihalSh

no problem!!!.........just to make it seem reasonable you can use $v=v_{0}+a.t$, solving for time gives $10~seconds$. It doesn't stop immediately, it slowly decelerates therefore time is comparatively large and so is the distance. I hope this makes more sense now!!!!

6. Oct 20, 2013

NihalSh

I got that, no problem.

7. Oct 20, 2013

Medgirl314

It does! Thanks, you have given some of the nicest replies I have found on here. If you have a moment, maybe you would go to my other Equations for Accelerated Motion problems and check those?

Thanks again!!

8. Oct 20, 2013

HA! :rofl: