# Distance Traveled

• Jan 18th 2011, 09:08 PM
PPSSAGW
Distance Traveled
Hi everyone. I'm working on a calculus problem about distance. I haven't learned it yet, but I have done some research to find a way to solve it.

A particle moves on the x-axis so that its velocity at any time t≥0 is given by $v(t)=12t^2-36t+15$

Find the distance the particle has traveled from t=0 to t=2.

Should I just calculate: $\int_0^2 (12t^2-36t+15) dt$ ? The answer for that is -10. Can distance be negative?

Thanks a lot.
• Jan 18th 2011, 10:04 PM
Unknown008
Well, to get the 'big picture', you can make a sketch.

$12t^2 - 36t + 15 = 3(4t^2 - 12t + 5) = 3(2t - 5)(2t - 1)$

(it's strange that it's very similar to http://www.mathhelpforum.com/math-he...tml#post605939)

You'll see that the graph a large part of the graph is negative between 0.5 and 2.

This means that the particle what initially going forward, but decelerating until changing direction. Then in accelerated, but within the 2nd second, it's still moving backwards. The integration that you gave does not give the distance, but the displacement, that is the final position of the particle, relative to the starting point.

To find the distance covered, you'll need to work out:

$\displaystyle \int^{0.5}_0 12t^2 - 36t + 15\ dt + \int^2_{0.5} -(12t^2 - 36t + 15)\ dt$