# Average velocity question

1. Oct 19, 2009

### noname1

I need to calculate the average velocity of

d= Et²+Ft+G over the interval [t1;t2]

I after resolving i get to this function

avg v = (E(t2)² + Ft2 - E(t1)² + Ft1) / (t2-t1)

my question is it possible to go any further than this?

2. Oct 19, 2009

### Staff: Mentor

Your average velocity has an error. The Ft1 term in the numerator should be -Ft1. After you fix that, you can simplify your expression.

avg v = (E(t2)² + Ft2 - E(t1)² - Ft1) / (t2-t1)
= [E(t22 - t12) +F(t2 - t1)]/(t2 - t1)

Now factor (t2 - t1) out of each term in the numerator, and you can cancel with the same factor in the denominator.

3. Oct 19, 2009

### noname1

yea i for got to distribute the - sign

would this be it?

[t2 - t1 (E(t2-t1) + F)]/ t2 - t1 and than the solution would be

(E(t2-t1) + F)

correct?

4. Oct 19, 2009

### Staff: Mentor

No. a2 - b2 = (a - b)(a + b).

5. Oct 19, 2009

### noname1

than like this

[E (t2-t1)(t2+t1) + F (t2-t1)] / t2-t1 =

than i can cancel the top (t2-t1) with the bottom giving

E (t2-t1) + F

6. Oct 19, 2009

### Staff: Mentor

No again.

Also, you should write your expression this way: [E (t2-t1)(t2+t1) + F (t2-t1)] / (t2-t1)

7. Oct 19, 2009

### noname1

crossed out the wrong one, i meant

[E (t2+t1) + F] correct?

Last edited: Oct 19, 2009
8. Oct 19, 2009

### Staff: Mentor

Yes, that's better.

9. Oct 19, 2009

### noname1

and than for instantaneous i would take the derivative which is

2et² + f correct?

10. Oct 19, 2009

### Staff: Mentor

Yes, but I would use the same letters as in the original problem.
v = 2Et2 + F

11. Oct 19, 2009

### noname1

thanks for your help, now just have to figure out my other thread