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

The total distance of the race is 200 m. From our calculations, the distance covered in the first 20 s is 180 m. Thus, the distance covered in the last 5 seconds is:

$\text{Distance in last 5 s} = 200 - 180 = 20 \, m.$

Since the sprinter decelerates uniformly during the last 5 seconds, we can use the equation of motion:

$s = ut + \frac{1}{2} a t^2,$
where:

• $s = 20 \, m$ (the distance covered),
• $u = 9 \, m/s$ (the speed at the beginning of the deceleration),
• $t = 5 \, s$.

Rearranging gives:

$20 = 9 \times 5 + \frac{1}{2} a \times 5^2.$

This simplifies to:

$20 = 45 + \frac{25a}{2} \Rightarrow \frac{25a}{2} = 20 - 45 = -25.$

Thus, solving for $a$ gives:

$\Rightarrow a = -2 \, m/s^2.$

Using the value of $u$ calculated previously:

$u = 6.2 \, m/s.$

Now using the equation of motion again:

$6.2 = 9 + a \times 5.$

Rearranging gives:

$a = \frac{6.2 - 9}{5} = \frac{-2.8}{5} = -0.56 \, m/s^2.$

The deceleration of the sprinter during the last 5 seconds of the race is therefore:

$a = -0.56 \, m/s^2.$