Equations for Accelerated Motion 4

AI Thread Summary
A motorcycle traveling at 30 m/s comes to a stop over 150 m, prompting a calculation of its acceleration using the equation v^2 = vinitial^2 + 2aΔx. The initial attempt yielded an acceleration of -4 m/s², which was later corrected to -3 m/s² after clarifying the calculations. The negative value indicates deceleration, confirming that acceleration acts in the opposite direction of the initial velocity. The discussion also highlighted the importance of understanding the time taken for deceleration, which was calculated to be approximately 10 seconds. Overall, the conversation emphasized the correct application of equations in physics problems.
Medgirl314
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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!
 
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Medgirl314 said:

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.
 
Medgirl314 said:
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!
 
Medgirl314 said:
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!
 
NihalSh said:
I got that, no problem.

HA! :smile:
 
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