I have the acceleration and the time elapsed. Say the acceleration is 5m/s^2. I need to find the speed at second 3. How would I go about doing that?
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Well, what do you know about the relation between speed and acceleration? – Gerry Myerson Sep 13 '17 at 02:23
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As you can probably guess I am taking physics I. I have checked every reference I have, and have found nothing concerning this matter. So, nothing I suppose. – Ramiro Rocha Sep 13 '17 at 02:24
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Just assume I know nothing – Ramiro Rocha Sep 13 '17 at 02:25
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Do you know what acceleration is? Just the definition? – Sep 13 '17 at 02:28
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OK, then: acceleration is the derivative, with respect to time, of the speed. So now the question is, what do you know about derivatives and antiderivatives? – Gerry Myerson Sep 13 '17 at 02:29
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Yes, of course. It is the rate at which speed increases. – Ramiro Rocha Sep 13 '17 at 02:29
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@gerrymyerson could you simplify the terms you are using – Ramiro Rocha Sep 13 '17 at 02:30
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Ok, so when you said that you know nothing about the relation between the two, that probably wasn't very true. Regardless: Something increases by 5 units a second for 3 seconds. What's the result? – Sep 13 '17 at 02:30
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@user296602 I'm an idiot. Thank you for the answer. Please post this as an answer so that it can be accepted. – Ramiro Rocha Sep 13 '17 at 02:32
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@RamiroRocha You can write this as an answer and post it yourself if you'd like; it would be good practice. – Sep 13 '17 at 02:34
1 Answers
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$acceleration(t)=\frac{\delta V(t)}{\delta t}$
$\implies V(t)=\int aceleration(s) ds= \int \frac{5m}{s^2} ds = \frac{-5m}{s}+C$
With C an initial condition?
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