Question

A particle moves on a straight line with velocity function v(t) = sin ωt cos2ωt. find its position function s = f(t) if f(0) = 0.

Answer 1

The correct answer for this question is this one:Hope this helps answer your question and have a nice day ahead.
v( t)= sin( ωt) cos^2 ( ωt) 

s(t) = ∫sin(ωt) cos^2 (ωt) dt 

let cos(ωt) = u 

-ωsin(ωt) dt = du 

∫sin(ωt) cos^2 (ωt) dt = -(1/ω)∫u^2 du = -(1/ω)u^3 / 3 + const 

s(t) = -(1/ω)cos^3(ωt) / 3 + const 

s(0) = 0 => const = 1/(3ω) 

s(t) = (1 - cos^3(ωt))/(3ω)

Answer 2

By integrating the velocity equation, we will see that the position equation is:

$f(t) = \frac{cos^3(\omega t) - 1}{3}$

How to get the position equation of the particle?

Here we know that the velocity of the particle is:

$v(t) = sin(\omega t)*cos^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$f(t) = \int\limits {sin(\omega t)*cos^2(\omega t)} \, dt$

If we take:

u = cos(ωt)

Then:

du = -sin(ωt)dt  ⇒  dt = -du/sin(ωt)

Replacing that in our integral we get:

$\int\limits {sin(\omega t)*cos^2(\omega t)} \, dt\\\\-\int\limits {\frac{sin(\omega t)*u^2 \, du}{sin(\omega t)}\\\\\\-\int\limits {u^2} \, dt = -\frac{u^3}{3} + c$

Where C is a constant of integration.

Now we remember that u = cos(ωt)

Then we have:

$f(t) = \frac{cos^3(\omega t)}{3} + C$

To find the value of C, we use the fact that f(0) = 0.

$f(t) = \frac{cos^3(\omega *0)}{3} + C = \frac{1}{3} + C = 0\\\\C = -1/3$

Then the position function is:

$f(t) = \frac{cos^3(\omega t) - 1}{3}$

If you want to learn more about motion equations, you can read:

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