A sprinter runs a race of 200 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1

To calculate the distance covered in the first 20 s, we can break it down into two parts:

1. Acceleration Phase (0 to 4 s):

   • Initial speed, ( u = 0 )
   • Final speed after 4 s, ( v = 9 \ m \ s^{-1} )
   • Time, ( t = 4 \ s )
   • Distance covered during acceleration:
   $$s_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \cdot a \cdot (4)^2$$

   To find acceleration, use:

   $$v = u + at \Rightarrow 9 = 0 + a \cdot 4 \Rightarrow a = \frac{9}{4} = 2.25 \ m \ s^{-2}$$

   Then, substituting back for distance:

   $$s_1 = \frac{1}{2} \cdot 2.25 \cdot 16 = 18 \ m$$

2. Constant Speed Phase (4 to 20 s):

   • Speed is constant at ( 9 \ m \ s^{-1} )
   • Time, ( t = 16 \ s )
   • Distance covered:
   $$s_2 = v t = 9 \cdot 16 = 144 \ m$$

3. Total Distance:

   $$s_{total} = s_1 + s_2 = 18 + 144 = 162 \ m$$

Thus, the distance covered in the first 20 s is 162 m.