Equations for Accelerated Motion 4

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SUMMARY

The discussion revolves around calculating the acceleration of a motorcycle that comes to a stop from an initial speed of 30 m/s over a distance of 150 m. The correct application of the equation v² = v_initial² + 2aΔx leads to the conclusion that the motorcycle's acceleration is -3 m/s², indicating deceleration. Participants clarified the calculations and addressed misconceptions regarding the interpretation of negative acceleration, emphasizing that it reflects a change in direction opposite to the initial velocity.

PREREQUISITES
  • Understanding of kinematic equations, specifically v² = v_initial² + 2aΔx
  • Basic knowledge of acceleration and deceleration concepts
  • Familiarity with units of measurement in physics (meters, seconds)
  • Ability to manipulate algebraic equations for solving variables
NEXT STEPS
  • Study the derivation and application of kinematic equations in various motion scenarios
  • Learn how to calculate time using the equation v = v_initial + a*t
  • Explore real-world applications of acceleration in automotive physics
  • Investigate the effects of different initial speeds and distances on deceleration rates
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Students studying physics, educators teaching kinematics, and anyone interested in understanding motion dynamics and acceleration calculations.

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.!
 
Last edited:
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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