Thread: Total distance? (Phys App of Calc)

1. Total distance? (Phys App of Calc)

I'm probably going to sound really stupid but

If a question asks "what is the total distance travelled in t seconds", how would one figure this out easily?

If needed I can find an example question, but I'd like a general solution to this problem

3. Re: Total distance? (Phys App of Calc)

Probably good idea to post an example question to.

4. Re: Total distance? (Phys App of Calc)

Omg you're a legend

Thank you!!

5. Re: Total distance? (Phys App of Calc)

It really depends what equation you're given, if you're given a displacement equation (the example form the khan academy link) it involves subbing points in as you can see.

However, if you're given a velocity equation, you have to find the area under the velocity curve using integration

e.g Your velocity equation is

$v = t^2-3t + 2$

You want to find the distance travelled in the first 2 seconds.

You compute

$\int_0^1 v dt + |\int_1^2 v dt|$

Since the area between t = 1 and 2 is below the curve (negative), and you need a positive value of that.

6. Re: Total distance? (Phys App of Calc)

Oh damn awesome

Thank you!!!

7. Re: Total distance? (Phys App of Calc)

Originally Posted by spaghettii
I'm probably going to sound really stupid but

If a question asks "what is the total distance travelled in t seconds", how would one figure this out easily?

If needed I can find an example question, but I'd like a general solution to this problem
$\noindent If given a velocity function v(t), the answer in general is \color{blue}\int_{t_{1}}^{t_{2}}|v(t)|\, dt. That is, to get the distance travelled over a given time interval, integrate the \textbf{absolute value} of v(t) over the time interval in question.$

$\noindent (By the way, the absolute value of v(t) is just the \underline{speed}, so the blue formula is just saying to integrate the speed over the time interval. This is just an extension of the formula distance = speed \times time'' to the case where the speed may not be constant.)$

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