Solving Uniform Acceleration: Calculating Velocity & Distance

AI Thread Summary
A user seeks help with a physics problem involving a car's uniform acceleration, needing to determine acceleration, initial and final velocities, and total distance traveled. They initially calculate the acceleration as 1.5 m/s² based on distance covered in two seconds but struggle with the reasoning behind their velocity estimates. Responses clarify that the acceleration must be derived from consistent kinematic relationships rather than direct distance differences. The discussion emphasizes the importance of correctly interpreting velocity and acceleration, highlighting a contradiction in the user's initial assumptions. Ultimately, the user is encouraged to reassess their calculations using proper kinematic equations.
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 \dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}
 
luitzen said:
The acceleration is \dfrac{dv}{dt}=\dfrac{\Delta v}{\Delta t}

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^2 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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