Average velocity: In a time of t seconds, a particle moves

[calculus 1983]

In a time of t seconds, a particle moves a distance of s meters from its starting point, where s = 4t^2 - t + 7

Find the average velocity of the particle between t = 5 and t = 5 + h if:

(a) h = 0.1, (b) h = 0.01, (c) h = 0.001

The formula for average velocity = s(b) - s(a) / b - a

Given: 4t^2 - t + 7

= s(5+h) - s(5) / (5+h) - 5

= 4(5+h)^2 - (5+h) + 7 - 4(5)^2 - (5) + 7 / h

when you have:
(5+h)(5+h) = 25 + 5h + 5h + h^2 = 25 + 10h + h^2 = 4(25 + 10h + h^2)
= 100 + 40h + 4h^2

when you have:
4(5)^2 = 4(5)(5) = 100


= 100 + 40h + 4h^2 - 5 - h + 7 - 100 + 5 - 7 / h

**On the denominator where we had (5+h) - 5 ... 5 and -5 cancelled each other out leaving us with h as the denominator

** 100 and -100 cancel out, -5 and 5 cancel out, and +7 and -7 cancel out

leaving us with = 40h + 4h^2 - h / h

when we factor the h out we have

h (40 + 4h - 1) / h

= 40 + 4h - 1

(a) h = 0.1
40 + 4(0.1) - 1 = 39.4 m/sec

(b) h = 0.01
40 + 4(0.01) - 1 = 39.04 m/sec

(c) h = 0.001
40 + 4(0.001) - 1 = 39.004 m/sec


Just double checking my work can anyone approve that this work is done correctly, thank you.

 

[soroban]

Re: Average velocity

Hello, calculus 1983!

Your work is correct . . . with a silly omission.

In \(\displaystyle t\) seconds, a particle moves \(\displaystyle s\) meters, where \(\displaystyle \,s(t)\:=\:4t^2\,-\,t\,+\,7\)

Find the average velocity of the particle between \(\displaystyle t\,=\,5\) and \(\displaystyle t\,=\,5\,+\,h\) if:

\(\displaystyle (a)\;h\,=\,0.1\;\;\;\;(b)\;h\,=\,0.01\;\;\;\;(c)\;h\,=\,0.001\)


The formula for average velocity: \(\displaystyle \:\frac{s(b)\,-\,s(a)}{b\,-\,a}\)

Given: \(\displaystyle \:s(t)\:=\:4t^2\,-\,t\,+\,7\)

\(\displaystyle \frac{s(5+h)\,-\,s(5)}{(5+h)\,-\,5} \:= \:\frac{\left[4(5+h)^2\,-\,(5+h)\,+\,7\right]\:-\:\left[4(5)^2\,-\,5\,+\,7\right]}{h}\)

. . \(\displaystyle = \:\frac{100\,+\,40h\,+\,4h^2\,-\,5\,-\,h\,+\,7 \:-\:100\,+\,5\,-\,7}{h}\)

. . \(\displaystyle = \:\frac{40h\,+\,4h^2\,-\,h}{h}\)

. . \(\displaystyle = \:\frac{h(40\,+\,4h\,-\,1)}{h}\)

. . \(\displaystyle = \:40\,+\,4h\,-\,1\;\) . . . Yes!

You did 99.9% of the required work . . . The rest is mere arithmetic.
. . Good work!

You overlooked: \(\displaystyle \:40\,+\,4h\,-\,1\:=\:39\,+\,4h\;\;\) *snicker*