How to Find Original Speed and Acceleration of a Truck Slowing Down

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Homework Help Overview

The problem involves a truck that covers a distance of 45.0 m in 8.50 s while decelerating to a final speed of 2.20 m/s. Participants are tasked with finding the truck's original speed and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various kinematic equations related to constant acceleration and attempt to apply them to the problem. Some express uncertainty about which formulas to use, while others suggest specific equations that might help in finding the original speed and acceleration.

Discussion Status

The discussion is active, with participants sharing their attempts and reasoning. Some have proposed formulas and calculations, while others question the validity of their approaches. There is no explicit consensus on the correct method or final values yet, but several participants are engaging with the problem and providing feedback on each other's reasoning.

Contextual Notes

Participants are navigating the constraints of the problem, including the need to find unknown variables without complete information. There is a focus on understanding the relationships between distance, speed, and acceleration in the context of the truck's motion.

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Homework Statement


A truck covers 45.0 m in 8.50 s while smoothly slowing down to a final speed of 2.20m/s
a. Find its original speed
b. find its acceleration


Homework Equations



average velocity=xf-xi/change in t
average acceleration=vf-vi/chang in t
final velocity=vi+at


The Attempt at a Solution



____________________________

0 8.5 s
2.20m/s

x-2.20m/s/(8.5-0s)

I know that the entire distance is 45.0 m. I don't think I can use any of the three formulas mentioned above, but I'm not sure which other one I could use. Could someone please show me what formula I need or give me a hint, please?

Thank you very much
 
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Since the truck is "smoothly slowing down", the acceleration must be constant and negative.

There are many nice formulas involving distance and speed with constant acceleration. Which ones do you know?
 
You're on the right track with some of the formulas you wrote. Think kinematics.
 
Thank you very much

I know that vf=vi+at, so vi must be=vf-at

acceleration=vf-vi/change in time

Therefore, vi=vf-[(vf-vi)/(change in time)*(t)] right? This doesn't seem to help.

I know that vf=2.20m/s and the change in time is 8.5. It seems that I need to know what vi or a is. 2.20m/s=vi+a(8.5) or would it just be v=d/change in time? vi=45/8.5=5.3
vf=vi+at
2.2=5.3+a(8.5)
a=-3.65

Could someone please explain to me how I could find vi, if I don't know what the acceleration is, please?

Thank you very much
 
Last edited:
You would just use this formula

d=\frac{V_i+V_f}{2}*t
 
Last edited:
Thank you very much

Does this look right?

d=vi+vft/2

45=vi+2.2m(8.50)/2

vi=71.3

a=vf-vi/t

a=2.2-71.3/8.5
=-8.13

Thank you
 
sorry didnt realize how the equation came out. You multiply time t by (V_f+V_i)/2
 
It's okay. Does this look right now?

45=8.5vi+18.7/2

vi=8.34

a=2.2-8.34/8.5=-.72

Thank you
 
Looks good to me, but I would get another person to check it just in case because I could be wrong.
 
  • #10
Thank you very much

Regards
 

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