in one hundred meter dash a sprinter accelerates from rest to a top speed with an acceleration 2.68?

Question

after achieving top speed he runs the remainder of the race without speeding up or down. if total race is run 12 seconds how far does he run during the acceleration phase

Answer 1

During a V-T run, person covers a total distance (=100m) in total time taken (T =12 s). The initial speed is zero (u=0 at t=0)
First he accelerated with f=2.68 m/s^2 achieved top speed V and covered S1 distance.
Thereafter the rest of distance S2 he just coasted along (with V).
Find S1??

During f:
S1 = 0+(1/2) f* t1^2
2 S1 = f* t1^2 ----(1)
(where t1 is time during accelerated phase)
V = f * t1 ---(2)

During coasting: [given >as T=t1+t2 =12 and S1 + S2 =100]
S2 = V * t2 = V (12 – t1)
100 – S1 = V (12 – t1)
S1 = 100 - V (12 – t1) -----(3)

(2) and (3) give
S1 = 100 – f * t1 (12 – t1)
S1 = 100 – 12 f * t1 + f t1^2 ----- (4) from (1)

S1 = 100 – 12 f * t1 + 2S1
S1 = 12 f * t1 – 100 put t1 = sqrt [2S1/f] from (1)

S1 + 100 = 12 f * sqrt [2S1/f]
squaring both sides
[(S1 + 100)]^2 = 144 f^2 * [2S1/f]
[(S1 + 100)]^2 = 288 f* S1 >>>> put
S1^2 + 200 S1 + 10000 = 288 f* S1 put f = 2.68
S1^2 - (571.84) S1 + 10000 = 0 solve for S1

S1 = (571.84) +- Sqrt [327000.9856 – 40000] / 2
S1 = [(571.84) +- (535.7247)] / 2

S1 (+) = 553.78 m (impossible – as race is just 100 m discarded)
S1 (-) = 18.06 m answer It means it just accelerated during his 18.06 meter stretch and took t1 = 3.67 s and achieved max V=9.835 m/s speed

Answer 2

d = a million/2 . 4 . t^2 = 2 t^2 v = 4 . t one hundred – d / (10 – t ) = v one hundred – 2t^2 = 4t ( 10 – t ) 2t^2 – 40t + one hundred = 0 clean up for t provides t = 2.ninety 3 and v accurate and very last % = 11.seventy 2 = 12 m/s