1. ## Antiderivatives

Hey,
I have an idea of how to approach this problem, but I dont understand how to solve it. Can someone help me please?

Problem:
A charged particle moves along the x-axis under the influence of an electric field. The field strength varies with time, and as a result, the velocity of the particle is complicated. The position of the particle at time t is written as x=x(t) and the velocity of the particle at time t is written as v(t).
Suppose we know that x(0)=0, and also that
v(t)={2t-1, if 0<t<1
{4t-3, if 1<t<2
{6t-7, if 2<t<3

What is x(1)? (...find x(2), x(3) and sketch x=x(t), v=v(t))

So, obviously we can take the antiderivative of v(t) to find x(t) from which you get: $t^2-t+C, 2*t^2-3*t+C, 3*t^2-7*t+C$ ..whats confusing me is the three different intervals for time t. how do i solve for x(1) if there are 2 different antiderivatives for it?? Im confused

2. Originally Posted by coe236
A charged particle moves along the x-axis under the influence of an electric field. The field strength varies with time, and as a result, the velocity of the particle is complicated. The position of the particle at time t is written as x=x(t) and the velocity of the particle at time t is written as v(t).
Suppose we know that x(0)=0, and also that
v(t)={2t-1, if 0<t<1
{4t-3, if 1<t<2
{6t-7, if 2<t<3

What is x(1)? (...find x(2), x(3) and sketch x=x(t), v=v(t))

So, obviously we can take the antiderivative of v(t) to find x(t) from which you get: $t^2-t+C, 2*t^2-3*t+C, 3*t^2-7*t+C$ ..whats confusing me is the three different intervals for time t. how do i solve for x(1) if there are 2 different antiderivatives for it?? Im confused
The function for x(t) is
$x = \int (2t - 1)~dt = t^2 - t + A$ for $0 \leq t < 1$

$x = \int (4t - 3)~dt = 2t^2 - 3t + B$ for $1 \leq t < 2$

$x = \int (6t - 7)~dt = 3t^2 - 7t + C$ for $2 \leq t < 3$

We know that x(0) = 0, so
$x(0) = 0^2 - 0 + A = 0 \implies A = 0$

And we know that x must be continuous at all times t. Thus
$\lim_{t \to 1}x(t) = 1^1 - 2 = 2 \cdot 1^2 - 3 \cdot 1 + B \implies B = 0$

and
$\lim_{t \to 2}x(t) = 2 \cdot 2^2 - 3 \cdot 2 = 3 \cdot 2^2 - 7 \cdot 2 + C \implies C = 4$

Thus
$x(t) = \left\{\begin{array}{rr} t^2 - t,\text{ if }0 \leq t < 1 \\
2t^2 - 3t,\text{ if }1 \leq t < 2\\
3t^2 - 7t + 4,\text{ if } 2 \leq t < 3\end{array}\right.$

-Dan