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JH
Numerade Educator

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Problem 65 Hard Difficulty

A particle moves on a straight line with velocity function $ v(t) = \sin \omega t \cos^2 \omega t $. Find its position function $ s = f(t) $ if $ f(0) = 0 $.

Answer

$$\frac{1}{3 w^{2}}\left(1-\cos ^{3} \omega t\right)$$

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Video Transcript

a particle moves on a straight line with velocity function V and we were to find it position function and were given an initial condition over here after zero zero. So we know that velocities the derivative of the position. So we know that FT or S they're they're equal is the anti derivative of the and now we can go ahead and replace fee with a sign of omega T times co sign squared omega T Now for this problem, we should go ahead and use the u substitution take you to be co sign Omega team So that do you is by the chain rule negative omega sign of a mega t titi and we can rewrite this negative to you over omega equals sign of omega T TNT. So we have Let's simplify this. So we have a negative one over Omega Integral You square, do you, which is negative? One over omega You cubed over three plus e Unless he's our combats. Are u substitution to rewrite? This is negative. Cho sang cute Omega team over three Omega plus e So again, this is just using our substitution And now we confined See because of our initial condition. Half of zero is zero. Oops. So let's use that. So we know zero well equal, Negative, cosign Cubed of zero over three omega pussy. And we are given that this is zero. So this means that sea is cosign cubed of zero over three Omega and we know Coastline of zero was one. So we have won over three Omega. So this is our value for sea. So our final answer using using this fags in our value for sea begin apply both disease at the same time. And we have ten s equals negative. Cho sang cute omega t over three Omega plus he which was won over three Omega. And that's our answer for us.