A runner of mass 51.5 kg starts from rest and accelerates with a constant acceleration of 1.35 m/s2 until she reaches a velocity of 5.7 m/s. She then continues running with this constant velocity. (Take the direction the runner is going to be the positive direction. Indicate the direction with the sign of your answer.)
3 answers
can't give an answer with no question. All you have done is present some data.
Oh, 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?
(a) How far has she run after 64.7 s?
(b) What is the velocity of the runner at this point?
v= 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.
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.
3 answers