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October 21, 2014

October 21, 2014

Posted by **James** on Friday, May 18, 2012 at 12:57pm.

- Physics - no question -
**Steve**, Friday, May 18, 2012 at 2:10pmcan't give an answer with no question. All you have done is present some data.

- Physics -
**James**, Friday, May 18, 2012 at 3:59pmOh, sorry I forgot to post the rest of the question.

(a) How far has she run after 64.7 s?

(b) What is the velocity of the runner at this point?

- Physics -
**Elena**, Friday, May 18, 2012 at 6:06pmv= a •t,

t=v/a =5.7/1.35 = 4.2 s.

The distance covered for this time is

s1 = a •t²/2 = 1.35 •(4.22)²/2 =12 m.

If 64.7 is the total time of the motion, then the time

of uniform motion is

t1= 64.7 -4.2 =60.5 s.

The distance for this time is

s2 = v •t1 = 5.7 •60.5 =344.9 m.

The total distance is s1+s2 = 12 +344.9 = 356.9 m.

If 64.7 sec. is the time of uniform motion, then

s2 = 5.7 •64.7 =368.8 m.

Therefore, the total distance is

368.8 + 12 =380.8 m.

The velocity at this point is 5.7 m/s as she runs at a constant speed.

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