Solving Uniform Acceleration: Calculating Velocity & Distance

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Homework Help Overview

The discussion revolves around a problem in kinematics involving uniform acceleration. The original poster presents a scenario where a car accelerates from rest, covering specific distances in consecutive time intervals, and seeks to determine the acceleration, velocities at different points, and total distance traveled.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the uniform acceleration and expresses confusion over their calculations and the velocity/time graph. Some participants discuss the relationship between acceleration and changes in velocity, while others question the assumptions made about the speed at the end of the first second.

Discussion Status

Participants are actively engaging with the problem, offering insights into the relationships between acceleration, velocity, and distance. There is a recognition of potential misunderstandings regarding the calculations and assumptions, particularly concerning the speed at the end of the first second.

Contextual Notes

The original poster mentions having difficulty with the problem despite drawing a velocity/time graph, indicating a possible gap in understanding the underlying concepts of uniform acceleration. There is also a suggestion that the original calculations may be incorrect, prompting further exploration of kinematic relationships.

Bmrboi
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Hi guys,

New here, i have a question which has been doing my head in for days now. Hope someone might know how to answer it.

The question is: A car travels with uniform acceleration from rest and is observed to cover distances of 6m and 7.5m respectively in two consecutive intervals of 1 second each.
Determine;

i) the value of the uniform acceleration,

ii) the velocities at the beginning and end of each of these two seconds,

iii) the total distance traveled from rest till end of the 7.5m distance period.


I can work out part ii) and iii) if i had acceleration, but i cannot seem to work it out.
I have drawn a velocity/time graph but it just confuses me even more.

From my calculations i work a= 1.5m/s2

Any help would be great!

Thanks Bmrboi
 
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The acceleration is [itex]\dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}[/itex]
 
luitzen said:
The acceleration is [itex]\dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}[/itex]

Thanks for the reply, are you meaning a= (7.5m-6m)/ (2s-1s)? Which Equals 1.5m/s2?
 
Yes. The acceleration is the slope of the velocity/time graph.

To be correct, the speed is not in m, but in m/s.

The first time interval is 6m in 1s or 6m/1s=6m/s. This will get more important when problems become less obvious and you can recognize mistakes by checking for wrong dimensions.
 
Yes you are right. I have answers now! thanks!
 
Bmrboi said:
Thanks for the reply, are you meaning a= (7.5m-6m)/ (2s-1s)? Which Equals 1.5m/s2?

No, this is not quite right. If your acceleration is constant then you can indeed use this formula, but how have you figured out that the speed at the end of the first second is 6 m/s? If you think for a second you'll realize that you have no basis for thinking that. It would only be true if it uniformly had a speed of 6 m/s for the first second, in which case it would have zero acceleration (which isn't true). You can also prove that 1.5 m/s[itex]^2[/itex] is wrong: that means that every second, the speed grows by 1.5 m/s, by definition. But your claim is that the speed is 6m/s after the end of one second. This is a contradiction.

So you'll have to try again, with a different kinematics relationship.
 

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