Author Topic: Average speed (Read 1405 times) Tweet Share

Studyinghard · « **on:** May 13, 2011, 07:04:57 pm »

A truck travels with a speed v(t) = $\frac{60} {(t + t In (t))}$ km/hr
Calculate the average speed of the truck over the time interval [1, 8].

Graphite · « **Reply #1 on:** May 13, 2011, 07:06:19 pm »

[v(

-v(1)]/(8-1)

Studyinghard · « **Reply #2 on:** May 13, 2011, 07:11:11 pm »

dont you have to anti diff it? im not sure how to anti diff this one which is why i posted it

Graphite · « **Reply #3 on:** May 13, 2011, 07:24:07 pm »

Pretty sure you can simply do what I did.

luken93 · « **Reply #4 on:** May 13, 2011, 07:31:05 pm »

Quote from: Graphite on May 13, 2011, 07:06:19 pm

[v(-v(1)]/(8-1)

Isn't that the average increase in speed then?

I would've said anti diffing it,

$\frac{1}{8-1}\int_1^8}{\frac{60}{t+tlog_e(t)}dt$

Graphite · « **Reply #5 on:** May 13, 2011, 07:37:42 pm »

Yes, sorry that would be incorrect as I interpreted it as an displacement vs time equation

Studyinghard · « **Reply #6 on:** May 13, 2011, 07:43:13 pm »

How do you get the anti diff of that :s

Thu Thu Train · « **Reply #7 on:** May 13, 2011, 07:57:04 pm »

I don't think you need to know how to anti differentiate a logarithm.

~~P.S. The integral of velocity is displacement. mmkay?~~
DISREGARD THAT.

Just for fun

$\frac{60}{8-1} \int_1^8 \frac{1}{t + t\ln(t)} \mathrm{d}t = \frac{60}{7} \left[ \ln(1 + \ln(t))\right]_1^8$

taiga · « **Reply #8 on:** May 13, 2011, 09:17:52 pm »

Was helping my mate with this earlier

let u = ln(t)

du/dt = 1/t

dt= du * t

t = e^u

sub it all in and you have it in terms of stuff that you can antidiff for! (t's should cancel out at some stage)

That said, I don't think the question asks for average speed, re-read it.

Studyinghard · « **Reply #9 on:** May 13, 2011, 09:25:29 pm »

Quote from: taiga on May 13, 2011, 09:17:52 pm

Was helping my mate with this earlier

let u = ln(t)

du/dt = 1/t

dt= du * t

t = e^u

sub it all in and you have it in terms of stuff that you can antidiff for! (t's should cancel out at some stage)

That said, I don't think the question asks for average speed, re-read it.

You're making me worry -.-

taiga · « **Reply #10 on:** May 13, 2011, 09:44:38 pm »

My bad, I remember now.

Find the solution to the integral to find the distance, and then divide by the time taken to travel that distance

Speed = Distance / Time

Studyinghard · « **Reply #11 on:** May 13, 2011, 09:54:17 pm »

can someone show working out , im getting stuck in the middle

Graphite · « **Reply #12 on:** May 14, 2011, 11:44:29 am »

Is this question from specialist or methods? If it is in methods I am sure you only need to know how to do this by calculator.

Studyinghard · « **Reply #13 on:** May 15, 2011, 08:57:31 pm »

Nah its not methods

BubbleWrapMan · « **Reply #14 on:** May 17, 2011, 06:00:32 pm »

$\frac{1}{7}\int_{1}^{8}\frac{60}{t+t\log_{e}(t)}dt = \frac{60}{7}\int_{1}^{8}(\frac{1}{t}\times \frac{1}{1+\log_{e}(t)})dt$

Let $u = \log_{e}(t)$

Then $\frac{du}{dt} = \frac{1}{t}$

And
When $t = 1, u = 0$
When $t = 8, u = \log_{e}8$

$\therefore \frac{60}{7}\int_{1}^{8}(\frac{1}{t}\times \frac{1}{1+\log_{e}(t)})dt=\frac{60}{7}\int_{1}^{8}(\frac{1}{1+u}\times \frac{du}{dt})dt$
$=\frac{60}{7}\int_{0}^{\log_{e}8}\frac{1}{1+u}du$
$=\frac{60}{7}(\log_{e}|1+\log_{e}8|-\log_{e}|1+0|)$
$=\frac{60\log_{e}(1+3\log_{e}2)}{7}$

*shrug*

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