1057where is it said that the velocity of that point mass on reaching back to the initial position is 0...
rather equate the two distances which will be equal....
A point mass starts moving in a straight line with constant acceleration.After time to the acceleration changes its sign,remaining the same in magnitude.Determine the time t from the beginning of motion in which the point mass returns to the initial position.
(According to me,
since the body is starting from rest,u=0ms-1
v=ato
This is also the initial velocity for the second part of motion...
thus,for second part of motion ,
v=u-at
0=ato-a(t-to)
from this i am getting t=2to....given correct answer is (3.414)to!!!!! )
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4 Answers
1057
158even then also...i am getting the same result
For first part of motion....
s=ato22
for second part of motion,
s=ato(t-to)-a(t-to)22
=2atto-3ato22-at22(after simplifying)
after equating
ato22=2atto-3ato22-at22
we get a quadratic equation....
t2-4tto+4to2=0 (after eliminating a)
thsi is a perfect square.....and solving for t we again get...t=2to...!!!
21velocity after time t0=at0
time taken by mass for coming to rest after this = t0
distance traveled in one direction = 12at2 + 12at2 =at2
time taken to travel same distance in opposite direction is t1
at02=12at12
t1=√2t0
total time= 2t0+√2t0= 3.414t0