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 calculated as follows:

1. Distance during acceleration (first 4 s): Using the formula for distance under uniform acceleration: d_1 = rac{1}{2} a t^2

   • Here, a = rac{5 ext{ m s}^{-1}}{4 ext{ s}} = 1.25 ext{ m s}^{-2}.
   • Therefore,
     d_1 = rac{1}{2} imes 1.25 imes (4^2) = 10 ext{ m}

2. Distance at constant speed (from 4 s to 64 s):

   • Speed = 5 m s⁻¹, Duration = 60 s.
   • Therefore, $d_2 = 5 imes 60 = 300 ext{ m}$

3. Total Distance for the first 64 s: $ext{Total distance} = d_1 + d_2 = 10 + 300 = 310 ext{ m}$

To find $V$, we use the total distance of the race and the distances calculated earlier:

1. The race distance is 400 m. The distance covered until the 64 s mark is 310 m. Hence, the distance remaining for the last 20 s is: $d_{remaining} = 400 - 310 = 90 ext{ m}$

2. The average speed during the deceleration phase can be calculated:

   • The average speed is given by: ext{Average speed} = rac{5 + V}{2}

3. Since this speed is maintained for 20 s, we can write: $d_{remaining} = ext{Average speed} imes ext{time}$ 90 = rac{5 + V}{2} imes 20

4. Solving for $V$: $90 = 10(5 + V)$ $9 = 5 + V$ $V = 4 ext{ m s}^{-1}$

We can find the deceleration using the formula:

1. Knowing the initial speed $u = 5 ext{ m s}^{-1}$ and final speed $V = 4 ext{ m s}^{-1}$ over a time period of 20 s: a = rac{V - u}{t}

   • Substituting values: a = rac{4 - 5}{20} = rac{-1}{20} = -0.05 ext{ m s}^{-2}

2. Thus, the deceleration of the girl in the final 20 s is (-0.05 ext{ m s}^{-2}).