A girl runs a 400 m race in a time of 84 s - Edexcel - A-Level Maths Mechanics - Question 4 - 2011 - Paper 1

To find the distance run in the first 64 seconds:

1. In the first 4 seconds of acceleration:

   • Using the formula for distance with constant acceleration, we find:

   d_1 = rac{1}{2} a t^2

   with $a = \frac{5 - 0}{4} = 1.25$ m/s².

   Therefore,

   $d_1 = \frac{1}{2} \times 1.25 \times 4^2 = 10 ext{ m}.$

2. From 4 to 64 seconds, she runs at 5 m/s for 60 seconds:

   $d_2 = v \times t = 5 \times 60 = 300 ext{ m}.$

3. Total distance:

   $d_{total} = d_1 + d_2 = 10 + 300 = 310 ext{ m}.$

For the last part of the race, we know the total distance and the distances covered:

1. The total distance for the race is 400 m. So, using the distance covered:

   $310 + d_3 = 400,$

   where $d_3$ is the distance covered during deceleration.

   Therefore, $d_3 = 400 - 310 = 90 ext{ m}.$

2. During deceleration, using the average speed while decelerating:

   $average \, speed = \frac{5 + V}{2}$

   and the time is 20 seconds:

   $90 = 20 \times \frac{5 + V}{2},$

   which simplifies to: $90 = 10(5 + V)$

   so,

   $$$$

ightarrow V = 4 ext{ m/s}.$$

To find the deceleration:

1. The initial speed at the start of deceleration is 5 m/s and the final speed is V m/s, which we previously found to be 4 m/s.

2. Using the formula for deceleration:

   $a = \frac{v_f - v_i}{t} = \frac{4 - 5}{20} = -0.05 ext{ m/s}^2.$

This indicates that the girl is decelerating at a rate of 0.05 m/s².