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,
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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?
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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
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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 ?
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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!
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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 ?
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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
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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
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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
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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!!
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