💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 87 Hard Difficulty

A panicle moves along a straight line with displacement $ s(t), $ velocity $ v(t), $ and acceleration $ a(t). $ Show that
$ a(t) = v(t) \frac {dv}{ds} $
Explain the difference between the meanings of the derivatives $ dv/dt $ and $ dv/ds. $


$v(t) \frac{d v}{d s}$

More Answers


You must be signed in to discuss.

Video Transcript

Our goal is to show that this equation is true and also to talk about the difference between DVD tea and DVDs. So whenever you have this kind of derivative notation, think about the units on the top and the units on the bottom. So the V stands for velocity. So is the rate of change of velocity and the T stands for time. So it's with respect to time, whereas the other one you still have the V on the top. So it's the rate of change of velocity, but the S on the bottom means position. So with respect to position now, it works out that if you multiply these DVDs times or if you multiply DVDs times d S d T, it's almost like the DS is cancelled almost like fraction problem. And you end up with DVD T so that will help us when we go back and do the first part of the problem. All right, so the acceleration is a derivative of velocity with respect to type, So acceleration is DVD T. And we just saw that we could write DVD T as DVDs times de s DT So DBT t is DVDs times de s DT. Well, what's de s? DT de s DT is the rate of change of position with respect to time. And we know that as the velocity. Okay, so all we would have to do then is rearranged the order of these terms and we have a of T equals B of tee times DVDs.