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!!!!! )
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....
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