Q#8 (Past Exam Paper – June 2015 Paper 12 Q9)
A sprinter runs a 100 m race in a straight line. He accelerates from the starting block at a constant acceleration of 2.5 m s-2 to reach his maximum speed of 10 m s-1. He maintains this speed until he crosses the finish line.
Which time does it take the sprinter to run the race?
A 4 s B 10 s C 12 s D 20 s
Solution:
Answer: C.
The motion consists of 2 parts:
1. motion with constant acceleration of 2.5 m s-2
2. motion with constant speed of 10 m s-1
Let the duration of the acceleration be t1.
v = u + at
Time t1 = (v – u) / a = (10 – 0) / 2.5 = 4 s
Let the distance travelled during the acceleration be s1.
Acceleration a = 2.5 m s-2
Initial speed u = 0 (at start)
Final speed v = 10 m s-1
We want to find distance s1. The equation that relates a, v, u and s is
v2 = u2 + 2as
Distance s = (v2 – u2) / 2a
Distance travelled s1 = (v2 - u2) / 2a = (102 – 0) / (2 × 2.5) = 20 m
The remaining distance of the 100 m race is covered at the constant speed of 10 m s-1.
Remaining distance = 100 – 20 = 80 m
Let the time taken to cover the remaining distance be t2.
Speed = distance / time
Time t2 = distance / speed = 80 / 10 = 8 s
Total time taken = 8 + 4 = 12 s
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