Calculating Cyclist Speed: 10 Secs and 5 Secs

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SUMMARY

The discussion centers on calculating the distance covered by a cyclist over a 10-second interval, given an average speed of 1.8 m/s and a top speed of 3.6 m/s. The cyclist's speed is analyzed under the assumption of constant acceleration, leading to the conclusion that the initial speed was zero. The distance covered in the first 10 seconds is calculated using the formula for the area of a triangle, specifically (1/2) * vmax * tmax, where vmax is the top speed and tmax is the total time.

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The cyclist got a average speed of 1,8m/s and a top speed of 3,6sek
The time is 10 sek.
How do I get to know how long will the cyclist cycle the last 5 sek of the 10 sek. Can i conclude that with my formula?

Please help me
 
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If you mean that the top speed is 2 times the average speed, then
(in first approximation of constant acceleration) the lowest speed was zero.
So the graph of velocity vs time is through the origin, straight-line.

How far did he go in the whole 10 seconds? (½ vmax tmax)
this is the area of the triangle "under the curve" v vs t .

How far would he go, in the last half of the race?
(OK, what would his average speed be during this half?)
 

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