<p>We haven't learned how to integrate functions in calc yet, but we're using a little bit of it in physics. Any help on this would be greatly appreciated!</p>
<p>Integrate the following acceleration functions into velocity and position functions:</p>
<p>1) a = 2</p>
<p>2) a = 0</p>
<p>3) a = -3</p>
<p>Sorry if these are mind-numbingly easy, but I just want to confirm my method with what I'm actually supposed to do. Thanks again!!</p>
<p>Since a = dv/dt, you need to solve v = integral( a over time), to get
v = at + u (where u=initial velocity, t=time) -------------- Eqn(1)</p>
<p>Similarly, since v = ds/dt, you need to solve s = integral( v over time), to get
s = (0.5 a)t^2 + ut (where s=distance) -------------- Eqn(2)</p>
<p>Plug in the different values of a into Eqns(1) and (2), to get your required answers.</p>
<p>Ahh I see.. I'm sorry, I forgot to write down the time intervals also.</p>
<p>1) At 10 seconds</p>
<p>2) At 5 seconds</p>
<p>3) At 5 seconds</p>
<p>Are my answer correct?</p>
<p>1) a = 2
v = (2)(10) = 20 m/s *Our teacher told us to ignore "c" the constant
s = (0.5)(2)(10^2) = 100 meters</p>
<p>2) a = 0
v = (0)(5) = 0 m/s
s = (0.5)(0)(5^2) = 0 meters</p>
<p>3) a = -3
v = (-3)(5) = -15 m/s
s = (0.5)(-3)(5^2) = -37.5 meters</p>
<p>I feel like I may have done something terribly wrong. Is this correct?</p>
<p>It looks fine providing all the initial conditions were zero - Vinitial must = 0 and hinitial (initial height) = 0. Since the problem asked for "functions" instead of an answer, ignoring the intial conditions is erroneous. However, maybe the teacher modified the problem to account for the class' lack of integration experience.</p>
<p>Sorry, I wasn't being clear before. I meant, integrate the functions into velocity and position functions and then solve for the velocities and positions at their respective times. I think I've figured out enough to get an understanding of what's going on here. I'll ask my teacher tomorrow, and hopefully she will fill me in on the rest. Thanks though!</p>