Mechanics> displacement of a particle moving on a st...

displacement of a particle moving on a straight line is given by x=16t-2t2.Distance travelled by the particle during the first 2sec. isS1 and during first 6sec . is S2 find 3S1/S2??

Hansraj Gyanendra Singh Rajawat , 8 Years ago

Grade 12

9 Answers

Vikas TU

Last Activity: 8 Years ago

put t =2 for S1

S1 = 16*2 – 2*4 = 32 – 8 = 24 meter.

Now check when is its velocity becomes zero.

dx/dt = v = 16 – 4t = 0

at t= 4 sec.

at t = 4

x = 64 – 32 = 32

and last 2 sec. it travelled zero distance,

thus at 6 s too we would count ditance as 32.

so S2 = 32

ratio

3S1/S2 = > 3*24/32 => 3*3/4 = 9/4

Hansraj Gyanendra Singh Rajawat

Last Activity: 8 Years ago

Sorry, the correct answer is 5. and I didinot understand why in last 2 sec. it travelled zero distance?????

Vikas TU

Last Activity: 8 Years ago

Because velocity becomes zero at t=4.

thus at 6 s too we would count ditance as 32.

Therfore, in last 2 sec. it travelled zero distance.

Hansraj Gyanendra Singh Rajawat

Last Activity: 8 Years ago

it will reverse its direction so ot will travel backward? it may also be possible? what do you think?? about this

Vikas TU

Last Activity: 8 Years ago

Yes!

for last 2 second i mean after 4 seconds it stops and then start moving back.

so,

Retardation, = -4 m/s^2

at t = 2

Therfore,

x = 0 + 0.5*4*4 = 8 meter.

Thus, total distance it travelled is: 32 + 8 = 40m =S2

ANd hence, 3S1/S2 = 3*24/40 = 3*6/10 = 3*3/5 = 9/5

BUT IT STILL IS 9/5 NOT 5.

Approved

Vivek Prakash

Last Activity: 7 Years ago

His logic is right and s1 = 24m and s2= 40m but at last there is a calculation mistake it will be (3*40)/24 = 5 answer

Harsh vardhan singh

Last Activity: 7 Years ago

On differentiating the given equation with respect to t we get,dx/dt(v)=16-4tand again differentiating we get acceleration(a)= -4Now the initial velocity I.e at t=0,will be(Initial)v=16We have to check when the displacement will equal to zero I.e particle will be at its initial position so putting x=o we get t=8 sec ,so at the 8 sec particle will be at initial position since the particle is moving on straight line so S1(distance travelled by the particle during the first two second) will come from the given equation by putting t=2 i.e S1=24Now if the particle`s displacement is zero at t=8sec we can conclude that it will move till 4sec and after that it will return back in the 5 sec, now the most important point here to consider is that displacement is time dependent which means that until 4th sec all the displacement will be different for each second but for the 5th sec it will be same as in the 4th sec and for the 6th sec it will be same as in the 3rd secNow for calculating S2 we have to calculate the distance travelled till the 4th sec and distance travelled in the fifth and sixth second, individuallyDistance travelled during first four second comes out to be 32 And now for calculating the distance travelled in 5th sec we have to calculate the distance travelled in 4th sec so by using formula, s=u+a(2n-1)/2 where u is initial velocity I.e 16 and a is acceleration I.e -4 and n=4sec we get the distance travelled in the 4th sec=2 and similarly in the 3rd sec we get the distance =6S2=32+2+6 I.e 40 3S2/S1=3*40/24 I.e 5 ans

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.

Rishi Sharma

Last Activity: 4 Years ago

Dear Student,
Please find below the solution to your problem.

For last 2 second, I mean after 4 seconds it stops and then start moving back.
so,
Retardation, = -4 m/s^2
at t = 2
Therfore,
x = 0 + 0.5*4*4 = 8 meter.
Thus, total distance it travelled is: 32 + 8 = 40m =S2

And hence, 3S1/S2 = 3*24/40 = 3*6/10 = 3*3/5 = 9/5
But it still is 9/5 not 5.

Thanks and Regards