A sprinter runs a race of 200 m. Her total time for running the race is 25 s. Figure 2 is a sketch of the speed-time graph for the motion of the sprinter. She starts from rest and accelerates uniformly to a speed of 9 m s⁻¹ in 4 s. The speed of 9 m s⁻¹ is maintained for 16 s and she then decelerates uniformly to a speed of u m s⁻¹ at the end of the race. Calculate (a) the distance covered by the sprinter in the first 20 s of the race, (b) the value of u, (c) the deceleration of the sprinter in the last 5 s of the race.

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

Question 3

A sprinter runs a race of 200 m. Her total time for running the race is 25 s. Figure 2 is a sketch of the speed-time graph for the motion of the sprinter. She starts... show full transcript

Worked Solution & Example Answer:A sprinter runs a race of 200 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1

Step 1

(a) the distance covered by the sprinter in the first 20 s of the race

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Answer

To calculate the distance covered in the first 20 seconds, we can break it down into two parts: the distance covered during acceleration (first 4 seconds) and the distance covered at constant speed (next 16 seconds).

Distance During Acceleration (first 4 seconds):
- Use the formula for distance when accelerating uniformly:
$d = ut + \frac{1}{2}at^2$

Here, initial speed (u) = 0, acceleration (a) = (\frac{9 - 0}{4} = 2.25 , m/s²), and time (t) = 4 seconds.

So,

$d = 0 \times 4 + \frac{1}{2} \times 2.25 \times 4^2 = 0 + \frac{1}{2} \times 2.25 \times 16 = 18 \, m$
Distance at Constant Speed (next 16 seconds):
- Speed = 9 m/s, time = 16 seconds:
$d = speed \times time = 9 \times 16 = 144 \, m$
Total Distance Covered in First 20 Seconds:

$Total\, Distance = 18 + 144 = 162 \, m$

Step 2

(b) the value of u

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Answer

After the first 20 seconds, the race is not yet finished. Therefore, in the last 5 seconds (total time 25 s), the runner decelerates from 9 m/s to u. The total distance covered during the last 5 seconds can be calculated as:

Given the total distance covered is 200 m:

$162 + \frac{1}{2}(9 + u) \cdot 5 = 200$

This simplifies to:

$162 + \frac{5}{2}(9 + u) = 200$

Rearranging gives:

$\frac{5}{2}(9 + u) = 200 - 162 = 38$

Solve for u:

$5(9 + u) = 76 \, \Rightarrow 9 + u = \frac{76}{5} = 15.2\, \Rightarrow u = 15.2 - 9 = 6.2 \, m/s$

Step 3

(c) the deceleration of the sprinter in the last 5 s of the race

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Answer

In the last 5 seconds, the sprinter starts at a speed of 9 m/s and decelerates to u (which we found to be 6.2 m/s). Therefore, the deceleration a can be calculated as:

Using the formula:

$a = \frac{v - u}{t}$

where v is final speed (6.2 m/s), initial speed is 9 m/s, and t is 5 s:

$a = \frac{6.2 - 9}{5} = \frac{-2.8}{5} = -0.56 \, m/s²$

Therefore, the deceleration is 0.56 m/s².

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

Worked Solution & Example Answer:A sprinter runs a race of 200 m - Edexcel - A-Level Maths Mechanics - Question 3 - 2005 - Paper 1

(a) the distance covered by the sprinter in the first 20 s of the race

(b) the value of u

(c) the deceleration of the sprinter in the last 5 s of the race

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