# Answer to Question #25558 in Mechanics | Relativity for jiajia

Question #25558

Boo radley wants to run a 1500m race with an average velocity of 6.00m/s. after running the first 500m, Boo notices that the elapsed time is 1 min 40 seconds. panicking he accelerates for "t" seconds at 0.20m/s^2 and then finishes the race at his new faster speed. he makes it just in time. how ling did he acclerate for?

Expert's answer

Boo Radley wants to run a 1500m race with an average velocity of 6.00m/s:

t& = 1500m / 6m/s = 250 s

after running the first 500m, Boo notices that the elapsed time is 1 min 40 seconds. Therefore, next 1500-500=1000 m he must run for 250-1m40s=150s

He runs with velocity

v0=500m/100s = 5m/s

Therefore:

v0=5m/s

T = 150 s

S = 1000 m

and he accelerates for& "t" seconds at 0.20m/s^2:

S = v0*t + a*t^2/2 + (v0+a*t)*(T-t) = v0*t + a*t^2/2 + v0*T + a*T*t - a*t^2 - v0*t =& v0*T + a*T*t - a*t^2/2

1000 = 5*150 + 30*t - 0.1*t^2

t^2 - 300t + 2500 = 0

t1 = 250 s >T

t2 = 50 s &

Answer: t=50 sec

t& = 1500m / 6m/s = 250 s

after running the first 500m, Boo notices that the elapsed time is 1 min 40 seconds. Therefore, next 1500-500=1000 m he must run for 250-1m40s=150s

He runs with velocity

v0=500m/100s = 5m/s

Therefore:

v0=5m/s

T = 150 s

S = 1000 m

and he accelerates for& "t" seconds at 0.20m/s^2:

S = v0*t + a*t^2/2 + (v0+a*t)*(T-t) = v0*t + a*t^2/2 + v0*T + a*T*t - a*t^2 - v0*t =& v0*T + a*T*t - a*t^2/2

1000 = 5*150 + 30*t - 0.1*t^2

t^2 - 300t + 2500 = 0

t1 = 250 s >T

t2 = 50 s &

Answer: t=50 sec

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