dv m = ~kv dt mg; v(0) = Uo
Added by Diego H.
Close
Step 1
First, we need to separate the variables by moving all the terms with v to one side and all the terms with t to the other side: dv/m - (k/m) v dt = -g dt Show more…
Show all steps
Your feedback will help us improve your experience
Kelan H and 61 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
At any instant the equation governing the motion is ( $\mathrm{k}=$ positive constant) (a) $\mathrm{kv}^{2}+\mathrm{mg}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$ (b) $\mathrm{mg}-\mathrm{kv}^{2}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$ (c) $-\mathrm{mg}-\mathrm{kv}^{2}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$ (d) $\mathrm{kv}^{2}-\mathrm{mg}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$
According to Charle's Law (a) $\left(\frac{\mathrm{dV}}{\mathrm{dT}}\right)_{\mathrm{P}}=\mathrm{k}$ (b) $\left(\frac{\mathrm{dT}}{\mathrm{dV}}\right)_{\mathrm{P}}=\mathrm{k}$ (c) $\left[\frac{1}{T}-\frac{\mathrm{V}}{\mathrm{T}^{2}}\right]=0$ (d) $\mathrm{V} \alpha \frac{1}{\mathrm{~T}}$
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Watch the video solution with this free unlock.
EMAIL
PASSWORD