## zeesbrat3 one year ago A particle moves on a line away from its initial position so that after t hours it is s = 2t2 +3t miles from its initial position. Find the average velocity of the particle over the interval [1, 4].

1. zeesbrat3

@Hero @nincompoop @abb0t @Whitemonsterbunny17 @freckles @ganeshie8 @Miracrown @vera_ewing @jim_thompson5910

2. zeesbrat3

any ideas?

3. zeesbrat3

4. ganeshie8

average rate of change (average velocity) between two points is same as the "slope" of the secant line connecting those two points

5. zeesbrat3

So find the derivative and set it equal to 0?

6. ganeshie8

You're given $s(t) = 2t^2 +3t$ For average velocity in the interval $$[1,4]$$, you simply find the slope between points $$(1,s(1))$$ and $$(4,s(4))$$ : $\dfrac{s(4)-s(1)}{4-1}$

7. zeesbrat3

mean value theorem..

8. ganeshie8

Easy, thats just the slope formula!

9. zeesbrat3

$s(4) = 2(4)^2 + 3(4) = 32 + 12 = 44$ $s(1) = 2 + 3 = 5$ $\frac{ 44 - 5 }{ 4 -1 } = \frac{ 39 }{ 3 } = 13$

10. ganeshie8

Looks good!

11. zeesbrat3

That's it?

12. ganeshie8

Yep.

13. zeesbrat3

Wow! Thank you, as always :)

14. ganeshie8

np

15. zeesbrat3

@ganeshie8 what would be the unit? miles/hour^2?

16. ganeshie8

how do you measure velocity ?

17. zeesbrat3

$\frac{ m }{ s^2 }$

18. ganeshie8

Really ?

19. zeesbrat3

I was actually good at physics lol

20. zeesbrat3

I would normally say that but this problem has hours and miles

21. zeesbrat3

i lied its just $\frac{ m }{ s }$

22. zeesbrat3

$\frac{ m }{ s^2 }$ is acceleration

23. ganeshie8

$\text{average rate of change}=\dfrac{s(4)-s(1)}{4-1}$ Notice that the units for top is $$miles$$ and the units for bottom is $$hours$$. so the unit for average rate of change is $$miles/hour$$

24. zeesbrat3

So, just switch from metric to american

25. ganeshie8

average velocity of the particle over the interval [1, 4] is 13 miles/hour

26. zeesbrat3

Okay, thank you :)