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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: aronno on September 03, 2009, 12:32:09 am

Title: Ex 13E Q9
Post by: aronno on September 03, 2009, 12:32:09 am: Essentials textbook (variable forces)

Ex13E Q9
i cant do it can someone help. please.

thanks
Title: Re: Ex 13E Q9
Post by: Damo17 on September 03, 2009, 01:17:15 pm: Quote from: aronno on September 03, 2009, 12:32:09 am
Essentials textbook (variable forces)

Ex13E Q9
i cant do it can someone help. please.

thanks

9a)
You know the variable forces for $0 \le t \le 5$ and $t>5$ . You also know the mass of $10$ so acceleration is given by:
$a=\frac{dv}{dt}=\frac{14-2t}{10}$ , $0 \le t \le 5$
$a=\frac{dv}{dt}= \frac{100t^{-2}}{10}$ , $t>5$

then to find velocity at $t=10$ :

$v=\int_0^5 (\frac{7}{5}-\frac{t}{5})dt + \int_5^{10} (10t^{-2})dt=(7-\frac{25}{10})+(-1+2)=5.5ms^{-1}$
Title: Re: Ex 13E Q9
Post by: jimmy999 on September 03, 2009, 09:30:53 pm: Funnily enough my friend asked me for help on this question today
I like Damo17's way of solving the question as it's much quicker than my method.

Another way to go about it is using (14-2t), find the equation for velocity, then find the velocity at t=5 and this will be the initial velocity for the other equation. Hence find velocity for (100t^-2) force, and then sub t=10 to find the velocity.

This method does take longer but whichever method allows you to understand what you're doing better then go with that. I myself prefer the above method over my own

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