Homework Question: The velocity (v) of an object is the rate of change of position (y(t)), and the acceleration is the rate of change of velocity. If the acceleration of an object falling vertically is constant and equal to -g, write a differential equation and initial conditions for the position of the object and solve it by integrating twice, if the object is stationary at a height of y meters at the starting time.

I ended up with y(t)=-1/2gt^2+kt+C (where k and c are arbitrary constants) y(0) = C = 0 y(t) = -1/2gt^2+kt

However, the solution just had y(t) = -1/2gt^2

Am I wrong because I kept two arbitrary constants in y(t)=-1/2gt^2+kt+C and didn't combine them to be y(t)=-1/2gt^2+ct ???

I have been getting confused with knowing what i can and cannot do with the constants after integrating. What if there are two? What if one is multiplied with the time function while the other isn't, can i still combine (like above)?

Thank you in advance!

  • $\begingroup$ The fact that the object is stationary at starting time enables you to rule out $k$ after the first integration (it's the initial velocity), just like you did with $C$ after the second integration. $\endgroup$ – Ronald Aug 20 '18 at 12:53
  • $\begingroup$ How do you know it's velocity? Not that I doubt you, but if its a non-familiar equations where I wouldn't recognize it, how would I know that this is velocity leading me to equate it to 0? $\endgroup$ – ndyson0 Aug 20 '18 at 13:03
  • 1
    $\begingroup$ Well, you integrate the (in this case) constant function that gives you the acceleration, the rate of change of velocity. After the integration you end up with a function that gives you the velocity at a given time. ($v_t = gt + v_0$). After a second integration you know the position at a given time ($x_t = \frac{1}{2}gt^2 + v_0t+ x_0$), as the velocity is the rate of change of position. By the choice your reference system, the $x_0 = C$ and the $v_0 = k$ can be influenced (and vanish in many cases). $\endgroup$ – Ronald Aug 20 '18 at 13:30
  • $\begingroup$ Thank you, Ronald! Very helpful :) $\endgroup$ – ndyson0 Aug 20 '18 at 13:45

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