a particle starts from rest and travels horizontally with a constant acceleration...at some point it hits a wall where its velocity and acceleration are reversed without any change in their magnitude..if the paticle returns to the initial point find the ratio of the time taken for the forward and reverse journey??

Question

unklerhaukus · Answer

so the velocity[m/s] is the the same magnitude before and after collision,

unklerhaukus · Answer

the distance is the same, hence the time is also the same

unklerhaukus · Answer

the ratio must be one

anonymous · Answer

the answer is root2 +1

unklerhaukus · Answer

doesn't that imply that the particle has slowed down?

anonymous · Answer

i don't think that it implies that the particle has slowed down because the particle is accelerating, so the velocity which it will posses at the time of collision would  be different from the initial velocity.

unklerhaukus · Answer

oh yeah is has a constant acceleration

anonymous · Answer

can someone solve fot it

ajprincess · Answer

for the forward motion
say the time taken is t1 and say the distance is s.
s=1/2at1^2
v=at1
in both the equations u=0 because it starts from rest.
for the reverse motion
say the time taken is t2.
the distance will be equal.
so
s=vt2+1/2at2^2
 =at1t2+1/2at2^2
1/2at1^2=(at1t2+1/2at2^2)
t1^2=2t1t2+t2^2
t1^2-2t1t1-t2^2=0
t1^2-2t1t2+t2^2=2t2^2
(t1-t2)^2=2t2^2
t1-t2=sqrt2t2
t1=sqrt2t2+t2
t1=t2(sqrt2+1)
t1/t2=sqrt2+1

ajprincess · Answer

is that k?