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].

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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].

Mathematics
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any ideas?
help please..

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average rate of change (average velocity) between two points is same as the "slope" of the secant line connecting those two points
So find the derivative and set it equal to 0?
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 \((4,s(4))\) : \[\dfrac{s(4)-s(1)}{4-1}\]
mean value theorem..
Easy, thats just the slope formula!
\[s(4) = 2(4)^2 + 3(4) = 32 + 12 = 44\] \[s(1) = 2 + 3 = 5\] \[\frac{ 44 - 5 }{ 4 -1 } = \frac{ 39 }{ 3 } = 13\]
Looks good!
That's it?
Yep.
Wow! Thank you, as always :)
np
@ganeshie8 what would be the unit? miles/hour^2?
how do you measure velocity ?
\[\frac{ m }{ s^2 }\]
Really ?
I was actually good at physics lol
I would normally say that but this problem has hours and miles
i lied its just \[\frac{ m }{ s }\]
\[\frac{ m }{ s^2 }\] is acceleration
\[\text{average rate of change}=\dfrac{s(4)-s(1)}{4-1}\] Notice that the units for top is \(miles\) and the units for bottom is \(hours\). so the unit for average rate of change is \(miles/hour\)
So, just switch from metric to american
average velocity of the particle over the interval [1, 4] is 13 miles/hour
Okay, thank you :)

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