Question

Find x(t) given the following conditions:

Answer 1

My attempt: \[\frac{ 1 }{ 2 }a_{0}+V_{0}t+X_{0}\]

Answer 2

Are you in an algebra course, a pre-calculus course or a calculus course?

That would affect the feedback I'd give you.

Answer 3

Physics (with calculus)

Answer 4

u can use integration to solve it as u do

Answer 5

Yes. Recall the following:

1. If you are given a constant acceleration, then \[v=\int\limits_{}^{}a *dt\]... which will be a function of time. You must allow for a constant of integration, which takes on the given value of the initial velocity.
2. Integrate the resulting velocity equation with respect to time to obtain an equation for position as a function of time; it must have a constant of integration whose value you obtain from the problem statement.

Any questions about this, so far?

Answer 6

yeah I used integration

Answer 7

oh ok

Answer 8

So, basically a(t) = a_0 , v(t)= a_0t + initial velocity, x(t)= 1/2a_0t^2+v_0t + initial position ?

Answer 9

yes, but they give you info to find the "constants"
for example,

\[ v(t) = a_0 t + c\]
and you are given \[ v(t_1) = v_1 \\ a_0 t_1 + c= v_1 \\
c= v_1 -a_0t_1\]
and you have
\[ v(t)= a_0 t + v_1 -a_0t_1\]

Answer 10

So the integral of v(t) would be:
\[\frac{ 1 }{ 2 }a_{0}t^2 + (v_{1}-a_{0}t_{1})t + C\]

??

Answer 11

yes, and that is x(t)
using \[ x(t_2)= x_2 \]
to find the constant

Answer 12

in other words
\[ x(t) = \frac{ 1 }{ 2 }a_{0}t^2 + (v_{1}-a_{0}t_{1})t + C \]
and at t= \(t_2\) we know
\[ \frac{ 1 }{ 2 }a_{0}t_2^2 + (v_{1}-a_{0}t_{1})t_2 + C = x_2 \]
so we can find C (which is a bit complicated ...)

Answer 13

Ok so I get \[x(t)=\frac{ 1 }{ 2 }a_{0}t^2+(V_{1}-a_{0}t_{1})t+x_{2}-\frac{ 1 }{ 2 }a_{0}(t_{2})^2-(V_{1}-a_{0}t_{1})t_{2}\]

Answer 14

lol

Answer 15

if we had numbers, that would simplify quite a bit.
But this problem seems to be an exercise in how to find the constants at times not t=0

Answer 16

Yeah

Answer 17

Thank you :D