Question

when you integrate a function like 2t^2. The answer is 6t^3 + c. What is c? the y intercept?

Answer 1

Its an unknown constant that MIGHT have been there. Because when you differentiate something like x^2 + 3, you get 2x. But if you integrate back, you would only get x^2. Well, whered the original + 3 go? How do we know if it ever existed? We put + c there because we want to assume there might have been a constant that disappeared in differentiation.

Answer 2

ok thx.. unfortunatly that doesnt help me solve the math question im doing now xD

Answer 3

What is it? Lol.

Answer 4

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position is x0 =1m at t0 =0s. At t =1s, what are the particle's (a) position, (b) velocity, and (c) acceleration?

I some how managed to find part b and c already lol though i dont understand how i figured out c xD all ik is that my answer was right. I want help with part a

Answer 5

Vx = 2t^2
Xo = 1m
To = 0s
those are subscript

Answer 6

the x in Vx, and the o in Xo and To

Answer 7

Well, velocity is the derivative of position and acceleration is the derivative of velocity. So if you want position you integrate, velocity just plug in condition, acceleration differentiate.

So you integrate and get 2t^3/3 + C
But the initial position x is 1 at t = 0. So we can say
1 = 2(0)^3/3 + C
C = 1. So the whole function is
x(t) = 2t^3/3 + 1
SO looks like at t = 1 you just have 5/3 m

Velocity just uses the function you already have, so
2(1)^2 = 2

Acceleration you take the derivative of and get 4t, so with t = 1, 4(1) = 4

But the main idea is if you have + C in your problem and youre given information, theres likely a way to solve for it. It gave me a condition for position, at t = 0, x = 1. This MUST be true, so when I integrate to get position, I can use that information to figure out what C is

Answer 8

oh ok, thx so much!
I was wondering if you could help explain to me a bit more about the relation between possition, velocity and acceleration?

Answer 9

Well, its mainly because derivatives are measures of change. Once you start changing your position (derivative), you get some sort of velocity. From there you can have a constant velocity, but once that starts changing (derivative), you have acceleration.

Answer 10

ok i get it :) I have a few more questions to complete which ive went through them but got stuck. do you mind going through a few with me?

Answer 11

Yeah, Ill see if I can get them xD

Answer 12

Determine the object's velocity at times t=0 s, 2 s, 4 s, 6 s, and 8 s.