Question

An athlet starting from stationary moves with an acceleration 2.2m/s^2 for 5 seconds, then for other 5 seconds

Answer 1

Answer:

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, $a=2.2$ 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:

$slope, a=\frac{AD}{OD}\\2.2=\frac{AD}{5}\\AD =2.2\times 5=11\textrm{ m/s}$

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, $A_{tri}=\frac{1}{2}\times OD\times AD=\frac{1}{2}\times 5\times 11=27.5$

Area of rectangle ABCD is, $A_{rec}=AB\times AD=5\times 11=55$

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

$d_{total}=27.5+55=82.5\textrm{ m}$