## anonymous one year ago A particle moves along the x-axis with velocity v(t) = t^2 − 4, with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = 3 seconds.

1. Astrophysics

For the distance you need to know the positive and negative values because if you integrate from 0 to 3 you just have the displacement.

2. anonymous

so how would you find those

3. Astrophysics

Well we know v(t) = t^2-4 =(t+2)(t-2), notice this gives us the negative value when v(t) is going in the opposite directions, so you will have to take absolute values of when it goes from $\left| \int\limits_{0}^{2} t^2-4 dt\right| + \left| \int\limits_{2}^{3} t^2-4dt \right|$ notice that distance is just $\int\limits_{0}^{3} |v(t)| dt$

4. Astrophysics

From 0<t<2 it's negative and 2<t<3 it's positive so we take the absolute value for that reason

5. anonymous

so the answer would be -3.0333?

6. Astrophysics

No, remember we are taking the absolute value

7. Astrophysics

That makes negative = +

8. anonymous

oh yes so it would be 7.6

9. Astrophysics

distance can't be negative, it's a scalar value

10. anonymous

thank you so much!

11. Astrophysics

Yes, that looks good $\frac{ 23 }{ 2 } \approx 7.7$

12. Astrophysics

So it worked out?

13. anonymous

yes it did and it made a lot of sense the way you explained it thank you