Equations for Accelerated Motion 4

In summary: Thanks! I will definitely check them out. Glad I could help!In summary, a motorcycle traveling at 30 m/s decelerates to a stop over a distance of 150 m, with an acceleration of -3 m/s^2. This may seem like a small acceleration, but it takes approximately 10 seconds for the motorcycle to come to a complete stop.
  • #1
Medgirl314
561
2

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!
 
Physics news on Phys.org
  • #2
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:
  • #3
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
The car?? I meant the motorcycle. Sorry. Blame it on lack of coffee syndrome.
 
  • #5
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!
 
  • #6
Medgirl314 said:
The car?? I meant the motorcycle. Sorry. Blame it on lack of coffee syndrome.

I got that, no problem.
 
  • #7
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
NihalSh said:
I got that, no problem.

HA! :rofl:
 

1. What is the equation for calculating acceleration?

The equation for calculating acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

2. How do you find the final velocity in an accelerated motion problem?

The final velocity can be found using the equation vf = vi + at, where vf is final velocity, vi is initial velocity, a is acceleration, and t is time.

3. Can you explain the difference between average acceleration and instantaneous acceleration?

Average acceleration is the overall change in velocity over a given period of time, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration is calculated using the change in velocity divided by the change in time, while instantaneous acceleration is calculated using the derivative of the velocity function.

4. How do you handle negative acceleration in equations for accelerated motion?

Negative acceleration indicates that the object is slowing down. In equations, it is represented by a negative value for the acceleration. You can use the same equations as you would for positive acceleration, but the final velocity will be smaller than the initial velocity.

5. Can you use equations for accelerated motion to calculate displacement?

Yes, you can use the equation x = xi + vit + 1/2at^2 to calculate displacement, where x is displacement, xi is initial position, vi is initial velocity, a is acceleration, and t is time.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
387
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
616
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
947
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
702
Back
Top