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I am very confused as to how to determine directions in problems. For example I was doing the following problem

A bomber releases a bomb while flying vertically upwards at a velocity of 1500 ft/sec at a height of 10,000 feet. a) How long after it is released will it take the bomb to reach the ground. b) Immediately after releasing the bomb the bomber flies away horizontally at the rate of 1200 feet/sec. How far away from the point at which the bomb strikes the ground is the plane when the bomb strikes?

So for this question I decided to choose the downward direction as positive for the distance traveled and so for that to happen I know that my acceleration needs also to be positive. In consequence I had $a= 32$, $v= 32t+1500$ and $s= 16t+1500$. Then I just equated $s(t)$ to $10 000 feet $ to see when would the distance traveled by the bomb would be $10,000 feet $ however this did not give me the correcte answer. I am not sure what I did wrong. Please let me know where I went wrong. Now for the second part I took $1200 feet/sec $ as a velocity and decided to integrate it to find the distance traveled. However in the answer they just multiply 1200 feet by 100 second to find its distance. Why is that?

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  • $\begingroup$ If the bomber is flying straight up (???) then the initial velocity of the bomb is $-1500$. There is nothing wrong with choosing down as positive, but one has to be consistent. But maybe there is something useful about always treating down as negative. $\endgroup$ – André Nicolas Jul 20 '16 at 1:15
  • $\begingroup$ Why would that be ? $\endgroup$ – samuel Jul 20 '16 at 1:17
  • $\begingroup$ When it is "dropped" it shares the current velocity of the plane. As to why down always negative, doing the same thing each time may help avoid error. $\endgroup$ – André Nicolas Jul 20 '16 at 1:18
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$$a=\frac{dv}{dt}=32 \\ \int_0^t \frac{dv}{dt}\ dt = \int_0^t 32\ dt \\ v(t)-v(0)=32t \\ v(t) = 32t+v(0)$$ But $v(0)$ is $1500\ ft/s$ in the negative direction (because the bomb is initially travelling upward with the plane and you chose downward as the positive direction). So $v(t) = 32t-1500$. Then integrate again to find $s(t)$.

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  • $\begingroup$ Ok I see Thank you. Now I get it . So because it is opposite to the direction I have chosen as positive it is negative. However what does v(t) = 32t-1500 mean on intuition level. Can i consider it as a net velocity equation for the object ? $\endgroup$ – samuel Jul 20 '16 at 1:28
  • $\begingroup$ It means that initially it is moving at 1500 ft/s in the negative direction but is linearly slowing down and will eventually start moving in the other direction. In this particular case, that means the bomb is heading upward but will slow down and then start falling in the other direction -- picking up speed (still linearly) as it does so. $\endgroup$ – user137731 Jul 20 '16 at 1:30
  • $\begingroup$ Okay yes I see. Thank you a lot for all your help. I definitely appreciate all the help. $\endgroup$ – samuel Jul 20 '16 at 1:35
  • $\begingroup$ No problem. :-) $\endgroup$ – user137731 Jul 20 '16 at 1:41
  • $\begingroup$ Ok np I just did. $\endgroup$ – samuel Jul 20 '16 at 1:43

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