Bhargabjyoti Nath
Last Activity: 6 Years ago
The basic formula used here is
S = ut + (1/2) at2
where
S=Displcement
u= initial velocity
t=time elapsed
a= accelaration
Now, Here,
x = 16t - 2t2
So,
dx/dt = 16 - 4t
i.e., Velocity, V= 16 -4t.
Therefore, at t=4,
Velocity, V=0
Again,
dV/dt = -4.
Thus, Accelaration, a= -4.
Now we are able to calcute our required dispacement.
Here,
S1= Displacement travelled after 1st 2 seconds;
S3 = Displacement travelled after 1st 3 seconds.
So,
S1 = ut + (1/2) at2.
->S1= 16×2 + (1/2)×(4)×(22) {Here, Accelaration is assumed to have positive direction)
->S1= 24
Again,
S3 = 16×4+(1/2)(-4)(16) + S3/
{ where S3/ is the distance travelled upto 2 seconds but in opposte direction i.e., in the direction where 'a' has positive direction}
->S3 = 32 + S3/
Now, I am explaining why I've taken the time to he 4 seconds. We see that the velocity is zero after 4 seconds. This gives us clue that after 4 seconds, the particle will have Opposite direction. So, the particle will have a distance aftee 4 seconds travelled as given below
S3= (Displacement upto 4 seconds in that direction where 'a' has negative direction) + (Dispacement upto 2 seconds in that direction where 'a' has positive direction).
Here,
S3/ = ut + (1/2)at2
Here,
u= velocity at t=4 seconds, i.e., 0
a= +4 m/s2(opposite direction)
t= 2 seconds.
Therefore,
S3/= 0 + (1/2)(4)(22)
->S3/ = 8 m
Thus,
S3 = 32 + 8 = 40 m
Therefore,
Required ratio is
( 3 × S3 ) ÷ (S1) = (3 ×40) ÷ (24) = 120 ÷ 24 = 5.