position and movement

Question

In a 100m race, a runner accelerates from rest at 3.3ms^-2 and reaches a

miximum velocity of 10ms^-1. She then maintains this velocity for the rest of the

race. What is her finishing time?

急唔該列算式同解釋一下，快考試plz!!!!!!

Update:

ans is 13.5m

Answer 1

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Time needed for reaching the MAXIMUM velocity :

u=0, v=10ms^-1, a=3.3ms^-2,

By v=u+at,10=0+3.3t

t=10/3.3 s

distance of the runner's velocity change from 0 to 10ms^-1 :

By v^2-u^2=2aS, 10^2-0=2(3.3)S

S=100/6.6 m

as the maitain the velocity when she reach a maximum velocity

so, the rest of the time =(100-100/6.6) /10 s

~8.48s

So,

Finishing time=10/3.3+8.48

~11.51 s

Source(s): myself

Answer 2

From t = 0 to t = 10/3.3 = 3.03 s, the displacement is:

s = 0 + 3.3 x 3.032/2 = 15.2 m

Then for the remaining 84.8 m, she needs 8.48 s to finish.

Thus total time needed = 8.48 + 3.03 = 11.51 s

Source(s): Myself