Removal of absolute signsAn object is dropped from a cliff. The object leaves with zero speed, and t seconds later its speed v metres per second satisfies the differential equationdv&nbsp;dt&nbsp;=10−0.1v2So I found t in terms of vt=12ln|10+v10−v|The questions goes on like this: Find the speed of the object after 1 second. Part of the answer key shows thist=12ln|10+v10−v|2t=ln|10+v10−v|e2t=10+v10−vSo here's my question: why is it not like this?±e2t=10+v10−vWhy can you ignore the absolute sign?

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Removal of absolute signsAn object is dropped from a cliff. The object leaves with zero speed, and t seconds later its speed v metres per second satisfies the differential equationdv&amp;nbsp;dt&amp;nbsp;=10−0.1v2So I found t in terms of vt=12ln|10+v10−v|The questions goes on like this: Find the speed of the object after 1 second. Part of the answer key shows thist=12ln|10+v10−v|2t=ln|10+v10−v|e2t=10+v10−vSo here&#039;s my question: why is it not like this?±e2t=10+v10−vWhy can you ignore the absolute sign?

Esteban Sloan · Accepted Answer

Answer:In general what you wrote is correct,t+c=ln|10−v10+v|has to be transformed to10−v10+v=±ecet=Cet,where&amp;nbsp;C≠0&amp;nbsp;in this context. However&amp;nbsp;C=0&amp;nbsp;and&amp;nbsp;C=∞&amp;nbsp;correspond to the previously excluded (for the separation approach) constant solutions.C is a constant over all the solution, its value and thus its sign gets determined by the initial condition. At&amp;nbsp;t0=0&amp;nbsp;this givesC=10−v010+v0

tabido8uvt · Accepted Answer

Step 1You don&#039;t ignore the sign. Let&#039;s look at the original equation, and notice what&#039;s happening at v=10. The acceleration is zero:dvdt=10−0.1⋅102=0Step 2Since you started with v=0, you can see that the object accelerates, initially with dvdt=10, but then the acceleration starts to decrease, as the object approaches v=10. After that the velocity can&#039;t increase any more. So with your initial condition, you have v&amp;lt;10. Then the sign of the expression in the absolute value is always positive. Note that if you would have started with an initial velocity greater than 10, then the acceleration would have been negative, but always v&amp;gt;10 (getting close to 10 as t→∞). Then you would use the negative sign when you explicitly write the absolute value.

Removal of absolute signs An object is dropped from

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