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

To find the value of T, we use the total distance formula:

1. The distance covered during the acceleration phase can be calculated using the formula for distance under constant acceleration:

   $ext{Distance}_{ ext{acc}} = \frac{1}{2} a t^2$

   Since the final speed reached is 8 m/s after 5 seconds:

   $a = \frac{8 - 0}{5} = 1.6 \text{ m/s}^2$

   Thus,

   $ext{Distance}_{ ext{acc}} = \frac{1}{2} \times 1.6 \times 5^2 = 20 \text{ m}$

2. The distance covered during the constant speed phase is:

   $ext{Distance}_{ ext{const}} = 8T$

3. The distance covered during the deceleration phase can be expressed in terms of the total time and distance:

   $500 = 20 + 8T + ext{Distance}_{ ext{decel}}$

   The total time taken is 75 seconds, so time during deceleration is:

   $t_{ ext{decel}} = 75 - (5 + T)$

4. Plugging in the values, and since the athlete comes to a stop, we can infer:

   \frac{1}{2} \times 8 \times (75) = 500$$

5. Solving the equation:

   $20 + 8T = 500$

   $8T = 480$

   $T = 60 \text{ seconds}$