Thread: Rectilinear Motion

The acceleration function and the initial velocity for a particle moving along a line are: a(t)= 2t + 3, v(0)= -4, 0<t<3

Find the distance traveled during the time interval


10, my answer, isn't correct. I tried subtracting v(0) from v(3), but it doesn't yield the correct answer. How should I approach this?

They want the distance traveled. v(t) is the velocity function. So evaluating v(3) - v(0) would not be correct.

Integrate a(t)
$\displaystyle \displaystyle \int 2t + 3 \,dt = t^2 + 3t + C = v(t)$

To find C, plug in the initial condition:
$\displaystyle v(0) = 0^2 + 3(0) + C = -4 \rightarrow C = -4$
So
$\displaystyle v(t) = t^2 + 3t - 4$

I'll call the position function s(t). Integrate v(t):
$\displaystyle \displaystyle \int t^2 + 3t - 4 \,dt = \dfrac{t^3}{3} + \dfrac{3t^2}{2} - 4t + C = s(t)$

Can you take it from here?


EDIT: Typos. TT_TT

$\displaystyle d(t) = \int v(t)~dt$

So integrate twice.

Originally Posted by eumyang

I'll call the position function s(t). Integrate v(t):
$\displaystyle \displaystyle \int 2t + 3 \,dt = \dfrac{t^3}{3} + \dfrac{3t^2}{2} - 4t + C = s(t)$

Little bit of a typo copy/paste mistake here..