Question

A particle moves at a velocity given by v(t)=6sin(t/2). If the particle is at x=1 when t=0, then its position when t=pi is x=?

Answer 1

do you know what the position function would be?

Answer 2

I think it would be the antiderivative?

Answer 3

Nope, it will be the integral of the velocity function. @Euler271 thanks for mentioning position function I forgot it at first and was pondering where does x come from?

Answer 4

The integral of the velocity function?

Answer 5

\[\int\limits_{a}^{b}v(t)dt\]

Answer 6

Something like that?

Answer 7

integral = anti-derivative

that is correct

Answer 8

tell me the indefinite integral of v(t) [no boundaries]

Answer 9

\[\int\limits_{?}^{?}6\sin(\frac{ t }{ 2 })\]

Answer 10

not quite. let u = t/2 du = dt/2

you get

\[\int\limits_{}^{}\sin \left(\frac{ t }{ 2 }\right)= -12\cos \left(\frac{ t }{ 2 }\right) + C\]

the + C needs to be added when you take the indefinite integral (meaning no boundaries)

Answer 11

i meant 6sin(t/2)dt

Answer 12

Oh. I wasn't quite sure what you meant. However, I did get that equation, too.

Answer 13

to find what C is for this particular scenario, you need to plug in the given value

x(t) = -12cos(t/2) + C

Answer 14

when t = 0, x(t) = 1

solve for C

Answer 15

sorry for long time to reply

Answer 16

So in essence its just 1=-12+C

Answer 17

yes

Answer 18

That's what I got.

Answer 19

so now x(t) = -12cos(t/2) + 13

you can now solve for x(t) for any given t. where x(t) is the position x at time t

Answer 20

Thank you so much :)

Answer 21

glad to help :)