An athlete runs along a straight road - Edexcel - A-Level Maths Mechanics - Question 2 - 2010 - Paper 1

To find the value of T, we need to set up an equation based on the total distance covered:

1. The distance covered during acceleration for 5 seconds is given by the formula: $ext{Distance} = \frac{1}{2} \times \text{acceleration} \times t^2$ Here, the athlete accelerates to 8 m/s in 5 seconds, therefore the acceleration is: $a = \frac{8 \, \text{m/s}}{5 \, ext{s}} = 1.6 \, \text{m/s}^2$ Thus, the distance covered during this phase is: $d_1 = \frac{1}{2} \times 1.6 \times 5^2 = 20 \, \text{m}$

2. The distance covered at constant speed for T seconds is: $d_2 = 8 \, \text{m/s} \times T$

3. The distance covered while decelerating must sum up with the previous distances to total 500 m. By the time she stops, the equations provide: $d_1 + d_2 + d_3 = 500 \, ext{m}$ The deceleration phase is also plotted with the distance: $d_3 = \frac{1}{2} \times 8 \times t_{decel}$ Setting the entire equation: $20 + 8T + \frac{1}{2} \times 8 \times (75 - (5 + T)) = 500$

4. Simplify the equation and solve for T: After substituting the values and solving, we find: $T = 50 \, ext{s}$