Hundred metre sprinter increases her speed from rest uniformly at the rate of 1 ms upto three quarters of the total run and covers the last quarter with uniform speed. How much time does she take to cover the first half and the second half of the run?
Answers
Answer:
Explanation:
=> Here, hundred metre sprinter increases her speed from rest uniformly at the rate of 1 ms upto three quarters of the total run and covers the last quarter with uniform speed.
=> Thus, Total distance = 100 m
For the 1st half run, distance = 50 m
u = 0 m/s
a = 1 m/s²
=> So, s = ut + 1/2 at²
50 = 0 + 1/2 * 1 * t² = >t = 10 sec
=> At the end of 50m:
v= u + at = 0 + 1 * 10 = 10 m/s.
=> For the second half of run, we calculate for s = 25 m first [uniformly accelerated motion]
Here,
u = 10 m/s,
a = 1 m/s²
=> Therefore, v² = u² + 2as
v²= (10)² + 2 * 1 * 25
v²= 100 + 50
v²= 150
v = √150
v = 12.247 m/s
=> For time t,
v = u + at
12.24 = 10 + t
t = 12. 24 - 10
t = 2.24 s
=> For the last 25 m:
t = distance / speed
= 25 / 12.24
= 2.04 sec
=> Thus, for the second half of the run, time taken = 2.24 + 2.04 = 4.28s.
Hence, she takes 10 sec to cover the first half and 4.28s to cover the second half of the run.