If the velocity function measured in m/s is v(t)=t^(2)-2t-8 for a particle moving along a line. Find a) the displacement and b) the total distance traveled by the particle during the time interval [1,6]

For a I got -10/3 but I'm not sure if it's right and for b I'm getting different answers the more times I do it. Thanks for the help :)

Question

If the velocity function measured in m/s is v(t)=t^(2)-2t-8 for a particle moving along a line. Find a) the displacement and b) the total distance traveled by the particle during the time interval [1,6]  

For a I got -10/3 but I'm not sure if it's right and for b I'm getting different answers the more times I do it. Thanks for the help :)

misty1212 · Answer

HI!!

misty1212 · Answer

the first one is the integral right? 
$$\int_1^6(t^2-2t-8)dt$$

misty1212 · Answer

yeah cause you got $-\frac{10}{3}$ and that is right

misty1212 · Answer

then for the total distance travelled you have to break the integral apart
integrate over the interval in which the velocity is negative (going backwards) and then over the interval over which it is positive, going forwards

misty1212 · Answer

fortunately this one factors easily as 
$$(t+2)(t-4)$$ so you can see it is 0 if $t=4$ negative from 1 to 4, positive from 4 to 6

misty1212 · Answer

$$-\int_1^4(t^2-2t-8)dt+\int _4^6 (t^2-2t-8)dt$$ will do it

ribhu · Answer

@misty1212  nice stuff

misty1212 · Answer

let me know when you get 
$$\frac{98}{3}$$ 
@ribhu  thanks !

anonymous · Answer

It took me a while, but I finally got it! Thank you so much for your help :)