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Physics 7 Online
OpenStudy (anonymous):

The figure shows the velocity graph of a particle moving along the x-axis (d=9.4, T=2.8). Its initial position is x0 = 3.4 m at t0 = 0 s. At t1 = 1.1 s,

OpenStudy (anonymous):

What is the particle's velocity? What is the particle's position?

OpenStudy (shubhamsrg):

Are you comfortable with integration ?

OpenStudy (anonymous):

i guess ill try

OpenStudy (anonymous):

how would i start?

OpenStudy (anonymous):

hello are you there?

OpenStudy (shubhamsrg):

v = mt when t=T, v=2d => 2d = m(T) =>m = 2d/T hence the relation between v and t is v = (2d/T)t you can now find velocity of particle at t=1.1 for the position, we know v= dx/dt hence dx = (2d/T)t dt integrate both sides, and use the fact that x=3.4 at t=0 to find the constant of integration. Hope this helps ?

OpenStudy (anonymous):

hold on a sec let me work it out real quick

OpenStudy (shubhamsrg):

Take your time.

OpenStudy (anonymous):

i got 7.38 for the velocity

OpenStudy (anonymous):

7.38 m/s velocity

OpenStudy (shubhamsrg):

Yep, seems correct.

OpenStudy (anonymous):

im kind of confused for the position can you explain it please

OpenStudy (shubhamsrg):

v= dx/dt right ?

OpenStudy (anonymous):

yes

OpenStudy (shubhamsrg):

so we have on that substitution, dx/dt = (2d/T)t ?

OpenStudy (anonymous):

yes i get that

OpenStudy (shubhamsrg):

=> dx = (2d/T) t dt

OpenStudy (anonymous):

so we solve for dx right?

OpenStudy (anonymous):

we know t=1.1 T=2.8 and d=9.4...

OpenStudy (shubhamsrg):

So you're not comfortable with integration!

OpenStudy (anonymous):

i guess not im kind of rusty on integration. Im sorry

OpenStudy (shubhamsrg):

hmm, guess you need to learn integration before attempting this problem.

OpenStudy (anonymous):

then can you help me with integration

OpenStudy (anonymous):

please

OpenStudy (shubhamsrg):

Leme try. 2d/T = 6.71 right ?

OpenStudy (anonymous):

correct

OpenStudy (shubhamsrg):

=> dx = 6.71t dt

OpenStudy (anonymous):

ok

OpenStudy (shubhamsrg):

Now we integrate both sides. Can you tell me integral of dx ?

OpenStudy (anonymous):

would i be the anti derivative so hold on give me a sec

OpenStudy (anonymous):

so 3.355 t^2

OpenStudy (shubhamsrg):

yep. so we have x= 3.56 t^2 + C where C is the constant of integration, clear so far ?

OpenStudy (anonymous):

yes its clear

OpenStudy (shubhamsrg):

We also know x=3.4 at t=0, use this to find C

OpenStudy (anonymous):

ok give me a sec to solve

OpenStudy (anonymous):

oh duh so c = 3.4

OpenStudy (shubhamsrg):

Yep, so x= 3.56 t^2 + 3.4 just plug in t=1.1 to find x at that time and congrats, you're done!

OpenStudy (anonymous):

i got 7.71m

OpenStudy (shubhamsrg):

And that is correct! Bingo! ^_^

OpenStudy (anonymous):

i plugged it in and it says its wrong

OpenStudy (shubhamsrg):

eh ? Try 7.7076 ?

OpenStudy (anonymous):

it said it was wrong.

OpenStudy (shubhamsrg):

I don't see how that is wrong. Approximately speaking, we have done it all correct so far.

OpenStudy (anonymous):

oh i see what we did wrong the antiderivative was 3.357 not 3.56 minor mistake.

OpenStudy (anonymous):

but thanks alot for helping me i guess i need to brush up on calculus. Thanks alot!!

OpenStudy (shubhamsrg):

hmm..

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