# Equations for Accelerated Motion 4

## Homework Statement

A motorcycle traveling at 30 m/s comes to a stop over a distance of 150 m. What was the motorcycle's acceleration?

## Homework Equations

v^2=vinitial+2aΔx

## 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!

## Homework Statement

A motorcycle traveling at 30 m/s comes to a stop over a distance of 150 m. What was the motorcycle's acceleration?

## Homework Equations

v^2=vinitial+2aΔx

## 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!
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.!!!

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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!

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

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!

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!!!!

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

I got that, no problem.

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!!

I got that, no problem.

HA! :rofl: