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

The distance can be found by considering two parts:

1. The distance during acceleration (0 to 4 s): $d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \cdot \frac{5}{4} \cdot 4^2 = 10 \text{ m}$
2. The distance at constant speed (4 to 64 s): $d_2 = v t = 5 \cdot (64 - 4) = 5 \cdot 60 = 300 \text{ m}$ Combining both distances gives us: $d = d_1 + d_2 = 10 + 300 = 310 \text{ m}$

To find V, we use the total distance of the race (400 m):

1. Distance covered before deceleration = 310 m.
2. Distance during deceleration (from 64 to 84 s): $d_3 = \frac{(5 + V)}{2} \cdot 20 = 20 \cdot \frac{(5 + V)}{2}$ Setting up the equation: $310 + \frac{(5 + V)}{2} \cdot 20 = 400$ Solving: $\frac{(5 + V)}{2} \cdot 20 = 90 \Rightarrow 5 + V = 9 \Rightarrow V = 4 \text{ m/s}$

To find the deceleration:

1. The girl starts at a speed of 5 m/s and ends at V = 4 m/s over 20 seconds.
2. Using the formula: $a = \frac{v_f - v_i}{t} = \frac{4 - 5}{20} = \frac{-1}{20} = -0.05 \text{ m/s}^2$ Thus, the deceleration is 0.05 m/s².