Question

Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time (t), find the body's position at time (t). a=16 cos4t, v(0)=2, s(0)=5 OR See attachment.

Answer 1

Integrate a twice and substitute with the given points each time to find the constants. Try to do it yourself first.

Answer 2

First anti-derivative,
V(0)= 4sin(4t)+C =2
V(0)= 4sin(4(0))+C=2
0+C=2
C=2

S(0)= -cos(4t)+C= 5
S(0)= -cos(4(0))+C=5
-1+C=5
C=6

Like this?

Answer 3

Almost! You have first \(v(t)=4\sin{t}+2\). Now integrate this to find \(s(t)\).

Answer 4

I hope I'm not confusing you. I'm trying to make it as easy as possible. :D

Answer 5

Lol, umm I thought the antiderivative of 16 cos4t is 4sin(4t)?

Answer 6

Yeah, you're right. Just a typo, sorry.

Answer 7

But don't forget the constant that you found to be 2, you have to integrate it.

Answer 8

It's cool, aaaah so you have to find the derivative of 4sin(4(0))+2?

Answer 9

the anti-derivative yeah of 4sin(4t)+2.

Answer 10

I get 2t-cos(4t)+C in that case

Answer 11

Yep, now find c.

Answer 12

2t-cos(4(0))+C=5
2t-1=5
2t=6
t=3

Answer 13

:o?

Answer 14

Wait, I messed up.
s(0)=2t-cos(4(0))+C=5
=2t-1+C=5
=2t+C=6
C=6-2t
C=-2t+6
soooo we get, 2t-cos(4t)-2t+6

Answer 15

The body's position at time t will be at 2t-cos(4t)-2t+6 .....?

Answer 16

Looks good, except for the -2t term. When solving for the constant, t=0 everywhere (not just in the cos term)
s(t)= -cos(4t)+2t+6

Answer 17

Aaa, okay. Awesome! Thanks. :D

Answer 18

You've got it! Awesome!