A sprinter reaches his maximum speed Vmax fromm the
rest in 2s with constant a, then he maintains that
speed and finishes the 100 meters in overall 10s. Vmax
speed is
Answers
Answer:
As 100 Yards = 91.44 Meters
Let the maximum speed is v m/s
=>By v = u + at
=>v = 2.5a ---------------(i)
Let the sprinter cover s meter in initial 2.5 sec
=>By v^2 = u^2 + 2as
=>(2.5a)^2 = 0 + 2 x a x s
=>s = 3.13a ------------(ii)
For the rest of the distance i.e. 91.44 - s he takes (9.6 - 2.5) = 7.1 sec
By s = vt
=>91.44 - s = 7.1v
=>91.44 - 3.13a = 7.1 x 2.5a
=>a = 4.38 m/s^2
By putting this value in (i) :-
=>v = 2.5 x 4.38 = 10.95 m/S
Explanation:
PLS MARK AS BRAINLIEST PLS MARK AS IT WAS REALLY A DIFFICULT QUESTION
Answer: The time taken to reach max. speed =
v−u/a
=
= 2.5 ms
= 2.5 ms −2
= 2.5 ms −2
= 2.5 ms −2 (10−0) ms
= 2.5 ms −2 (10−0) ms −1
= 2.5 ms −2 (10−0) ms −1
= 2.5 ms −2 (10−0) ms −1
= 2.5 ms −2 (10−0) ms −1
= 2.5 ms −2 (10−0) ms −1 =4 s
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s =
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 =
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 m
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m=
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1 (100−20) m
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1 (100−20) m
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1 (100−20) m
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1 (100−20) m =8 s
= 2.5 ms −2 (10−0) ms −1 =4 sSo, total distance covered in this 4 s = 21 at 2 = 21 ×2.5×4 2 = 20 mTime taken to complete rest (100−20) m= 10 ms −1 (100−20) m =8 sSo, total time taken = (8+4)s=12 s
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