An athlet starting from stationary moves with an acceleration 2.2m/s^2 for 5 seconds, then for other 5 seconds
continues with steady speed up to the finish line.
-arrange in a graph speed due to time.
-what is the athlet's speed when crossing the finish line?
-calculate the distance covered from start till the finish line.

The graph is shown below.

The athlete's speed at the finish line is 11 m/s

Total distance covered is 82.5 m.

Explanation:

The graph of speed versus time is drawn below.

Acceleration in the first 5 seconds, m/s²

From the graph, the slope of line OA is the acceleration in the first 5 seconds.

Slope of line OA is given as:

Now, the length AD is nothing but speed at point A or B as AB is a straight line.

Therefore, the speed when crossing the finish line is the speed at B which is equal to 11 m/s.

Distance covered is given by the total area under the graph.

The total area can be divided into two shapes; a triangle and a rectangle.

The area under the graph is the sum of areas of triangle OAD and rectangle ABCD.

Area of triangle OAD is,

Area of rectangle ABCD is,

Therefore, the total distance covered till the finish line is given as:

friction

explanation:

they us to move forward and backwards

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