1.3 Iridov - Find Time Interval ( Kinematics )

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SUMMARY

The discussion focuses on a kinematics problem involving a car's motion characterized by acceleration, uniform velocity, and deceleration. The car accelerates at 5 m/s², travels for a total of 25 seconds, and achieves an average velocity of 72 km/h. The main question is to determine the duration of uniform motion, with participants suggesting the use of trapezoidal area calculations to solve the problem. The original poster encountered an imaginary result in their calculations, indicating a potential error in their approach to integrating the velocity function.

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razored
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Thank you for taking the time to read this lengthy post.

A car starts moving rectilinearly, first with acceleration a=5m/s2(the initial velocity is 0) then uniformly, and finally, decelerating at the same rate a, comes to stop.The total time of motion equals 25 seconds. The average velocity during whole time is 72 km/h. How long does the car move uniformly?WARNING : 2 MB IMAGE

http://img26.imageshack.us/img26/1403/defaultx.png" All my equations appear to be correct; however, when I calculate the t_{o}, I get an imaginary answer. What did I miss or do wrong?
 
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razored said:
Thank you for taking the time to read this lengthy post.

A car starts moving rectilinearly, first with acceleration a=5m/s2(the initial velocity is 0) then uniformly, and finally, decelerating at the same rate a, comes to stop.The total time of motion equals 25 seconds. The average velocity during whole time is 72 km/h. How long does the car move uniformly?

Maybe consider the area of a trapezoid?

That is what your V(t) graph represents.

You know that the integral of V(t) yields X(t)

They tell you what the area is ... avg V * total T = 20 * 25 = 500.

The base is 25 sec and total X is 500
So ...

(25+X)/2*H = 500

You also know that X = 25 - 2d
where d is the 2 periods of acceleration and deceleration.

And H = 5*d
where V = 5*t so ...

H = 5*(25 -X)/2

Combining things then yields:

5*(25 -X)/2*(25+X)/2 = 500

5/4*(252 - X2) = 500

I get a real result.
 
Actually, I ignored the graph I drew when I solved the problem because I realized(on another piece of paper) that the area I was calculating (two triangles and rectangle) was identical to D = 2x_o + x_1.

This may be asking too much but can you find the mistake in my work? I just don't see it.

Thanks again.
 
razored said:
Actually, I ignored the graph I drew when I solved the problem because I realized(on another piece of paper) that the area I was calculating (two triangles and rectangle) was identical to D = 2x_o + x_1.

This may be asking too much but can you find the mistake in my work? I just don't see it.

Thanks again.

Now that you know the answer, you can work backwards to figure the error in your math. In all honesty, I'm not that interested in looking.
 
EDIT : My mistake.
 
Last edited:

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