Question 8
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
Reference: Past Exam Paper – June 2015 Paper 12 Q9
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.
Distance travelled s1 = (v2 - u2)
/ 2a = (102 – 0) / (2 × 2.5) = 20 m
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
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
Thanks.
ReplyDeletethank you vey much please include small lines explaning why we used this formula etc
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