anonymous
  • anonymous
Find the velocity, v(t), for an object moving along the x-axis if the acceleration, a(t), is a(t) = 2t + sin(t) and v(0) = 4.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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alekos
  • alekos
a(t) = dv/dt so v(t) = int(2t+sint)
phi
  • phi
replace a(t) with dv/dt and write a= 2t+sin t as dv/dt = 2t + sin t dv = (2t +sin t ) dt
phi
  • phi
now integrate both sides

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phi
  • phi
can you do the integration ?
anonymous
  • anonymous
Kind of lost on what you're doing
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
|dw:1446429206231:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446429240051:dw| @Ephemera hopefully you agree that a(t) = dv/dt ?
anonymous
  • anonymous
Yeah
jim_thompson5910
  • jim_thompson5910
multiply both sides by dt |dw:1446429315716:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1446429328342:dw|
jim_thompson5910
  • jim_thompson5910
so we now have \[\Large dv = (2t+\sin(t))dt\]
jim_thompson5910
  • jim_thompson5910
now integrate both sides \[\Large dv = (2t+\sin(t))dt\] \[\Large \int dv = \int(2t+\sin(t))dt\] \[\Large \int 1 dv = \int(2t+\sin(t))dt\] \[\Large \int 1 dv = \int(2t)dt+\int(\sin(t))dt\] \[\Large v = t^2-\cos(t)+C\]
jim_thompson5910
  • jim_thompson5910
We're given `v(0) = 4` so that means `t = 0 and v(t) = 4` \[\Large v = t^2-\cos(t)+C\] \[\Large v(t) = t^2-\cos(t)+C\] \[\Large 4 = (0)^2-\cos(0)+C\] what is C equal to?
jim_thompson5910
  • jim_thompson5910
any ideas @Ephemera ?
anonymous
  • anonymous
No
anonymous
  • anonymous
I got a problem with different numbers so if you could finish this one up so I can apply the same method, I'd appreciate it.
jim_thompson5910
  • jim_thompson5910
what is `cos(0)` equal to?
anonymous
  • anonymous
1
jim_thompson5910
  • jim_thompson5910
so \[\Large 4 = (0)^2-\cos(0)+C\] \[\Large 4 = 0-1+C\] \[\Large 4 = -1+C\] \[\Large 4+1 = -1+C+1\] \[\Large 5 = C\] \[\Large C = 5\] agreed?
alekos
  • alekos
you gave him the answer and he just disappeared!! no thank you and no nothing. he didn't even attempt to try and solve any part of the problem. thats gratitude for you!!!
anonymous
  • anonymous
@alekos How about you mind your own business? I haven't even finished up my assignment and didn't have the time to check the progress made on the question. I was visiting my grandfather who has non treatable brain cancer. Your response is so unneeded, either delete it or I will be reporting it. And @jim_thompson5910 has helped me out multiple times so I am sure he is well aware that I appreciate his help, I always attempt to solve instead of just getting the answer.
alekos
  • alekos
I still haven't seen a thankyou to Jim who spent all that time and trouble to help you out

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