A sprinter runs a 100 metre race. He does not move for 0.2 seconds after the starting pistol is fired. Then his velocity vms-1 increases as v(t) = -0.5t2 + 5t - 0.98 where t is time in seconds. he reaches his maximum speed and maintains this until he crosses the finish line.
Calculate his time to complete the race.
when velocity is maximum
dv/dt = 0
-0.5*2*t + 5 - 0 = 0
-1*t = -5
t = 5
distance travelled in the first 5 s,
d1 = integral v*dt ( t = 0 to t = 5s)
= integral (-0.5*t^2 + 5*t + 0.98)*dt
= -0.5*t^3/3 + 5*t^2/2 + 0.98*t
= -0.5*(5^3 - 0^3)/3 + 5*(5^2 - 0^2)/2 + 0.98*(5 - 0)
= 46.57 m
so, velocity after 5 s, v = -0.5*5^2 + 5*5 + 0.9
= 13.4 m/s
remaining distance, d2 = 100 - d1
= 100 - 46.57
= 53.43 m
reaining time taken = d2/v_max
= 53.43/13.4
= 3.987 s
so, total time taken = 0.2 + 5 + 3.987
= 9.187 s <<<<<<<--------------------Answer
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