displacement

J

James Smithson

Junior Member
Joined
Nov 6, 2020
Messages
55
Hello everbody thank you for helping me in advance.

I have an issue in my book I believe i need to use intergration to solve the problem hence I am using the calculus forum.


My issue is I dont even know where t begin with this question. I need to learn how to solve problems like this but this is the only problem of its kind in my book and the explanation confused me (probably as english is a second language) please could someone really dumb it down for me so i can solve further problems like thisin the future.

thanks again :)
James

1726825957097.png
 
I'll try to walk you through it.

You know that \(\displaystyle v = \frac{ds}{dt} \). That is, velocity is the change in displacement with respect to time. Right?
 
sorry im not good at putting in the maths on here but i think i have to find the anti derivative of 4sin (3t) (ignore the constant)

is this -4cos(3t) and plug in the 0 then plug in pi/2 and add the difference together ? i really am clueless here haha
 
I'm going to try to help you UNDERSTAND what you are doing. Please answer my question.

You know that \(\displaystyle v = \frac{ds}{dt}\). That is, velocity is the change in displacement with respect to time. Right?
 
Ok so yes you have to integrate the velocity function to get displacement.

You have tried this and have asked whether the integral of 4sin(3t) is -4cos(3t).

If that is correct then the derivative of -4cos(3t) should be 4sin(3t). Is it?
 
James Smithson said:
sorry im not good at putting in the maths on here but i think i have to find the anti derivative of 4sin (3t) (ignore the constant)

is this -4cos(3t) and plug in the 0 then plug in pi/2 and add the difference together ? i really am clueless here haha
Click to expand...
What expression do you get when you differentiate [-4cos(3t)] - w.r.t. "t" - please show your work.
BTW, you can take a picture of your work and attach it with your post.
 
Last edited:
Harry_the_cat said:
Ok so yes you have to integrate the velocity function to get displacement.

You have tried this and have asked whether the integral of 4sin(3t) is -4cos(3t).

If that is correct then the derivative of -4cos(3t) should be 4sin(3t). Is it?
Click to expand...
- get 12 sin (3t) oops haha
 
Right! You can always check your integration by finding the derivative!
So have another go at finding the integral of 4sin(3t). Check your own answer before you reply with your result.
 
James Smithson said:
v=0 when t=0 subbing these values into the formular for v gives c=0 ? i think
Click to expand...
No that's not the reason. You don't know that v=0 when t=0. The object could already be moving when the timing starts. If you were told that the object moved "from rest" you could assume v=0 when t=0.

So let's see what happens if we DON'T ignore the constant.

Then \(\displaystyle s(t) =-\frac{4}{3} cos (3t) + c \).

So when \(\displaystyle t=0\), \(\displaystyle s =-\frac{4}{3} cos (3*0) +c = -\frac{4}{3} cos(0) + c = -\frac{4}{3} + c \) . You get that?
 
Harry_the_cat said:
No that's not the reason. You don't know that v=0 when t=0. The object could already be moving when the timing starts. If you were told that the object moved "from rest" you could assume v=0 when t=0.

So let's see what happens if we DON'T ignore the constant.

Then \(\displaystyle s(t) =-\frac{4}{3} cos (3t) + c \).

So when \(\displaystyle t=0\), \(\displaystyle s =-\frac{4}{3} cos (3*0) +c = -\frac{4}{3} cos(0) + c = -\frac{4}{3} + c \) . You get that?
Click to expand...
that makes more sense . thank you Harry the cat
 
When \(\displaystyle t=\frac{\pi}{2}, s = -\frac{4}{3} cos(\frac{3\pi}{2}) + c = 0 + c = c \).

To finish off, the CHANGE in displacement is \(\displaystyle s(\frac{\pi}{2}) - s(0) = c - (-\frac{4}{3} + c) = \frac{4}{3}\)

See how the c's cancel each other out. THAT'S why you can ignore them. They are not necessarily 0, but they will always cancel out.
 
You must log in or register to reply here.
Top