Velocity: A particle moves along the y-axis with velocity...

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fdragon

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A particle moves along the y-axis with velocity given by v(t)= tsin(t^2) for t >= 0.

A. In which direction (up or down) is the particle moving at time t = 1.5? Why?

B. Find the acceleration of the particle at time t = 1.5. Is the velocity of the paricle increasing at t = 1.5? Why or why not?

Don't know where to begin!
 
Re: Velocity?

Hello, fdragon!

Have they taught you nothing about velocity and acceleration ??

A particle moves along the y-axis with velocity given by \(\displaystyle v(t)\:= \:t\sin(t^2)\) for \(\displaystyle t\,\geq\,0\)

A. In which direction (up or down) is the particle moving at time \(\displaystyle t\,=\,1.5\)? .Why?
Click to expand...

When \(\displaystyle t\,=\,1.5\), we have: \(\displaystyle \:v \;=\;(1.5)\sin(1.5^2) \:\approx\:+1.167\)

Since the velocity is positive, the particle is moving upward.


B. Find the acceleration of the particle at time \(\displaystyle t\,=\,1.5\)
Is the velocity of the paricle increasing at \(\displaystyle t\,=\,1.5\)? .Why or why not?
Click to expand...

Acceleration is the derivative of the velocity.

Hence: \(\displaystyle \:a(t) \;=\;t\cos(t^2)\cdot2t\,+\,\sin(t^2) \;=\;2t^2\cos(t^2)\,+\,\sin(t^2)\)

When \(\displaystyle t\,=\,1.5:\;a(1.5) \;=\;2(1.5^2)\cos(1.5^2)\,+\,\sin(1.5^2) \;\approx\;-2.048\)

Since the acceleration is negative, the velocity is decreasing.

 
You're really awesome.... thank you so much.. now that I think about it I could've done the first part but the second part was what I really needed the help on
Thanks again!

Ok so why did you add sin(t^2) at the end?
 
fdragon said:
You're really awesome.... thank you so much.. now that I think about it I could've done the first part but the second part was what I really needed the help on
Thanks again!

Ok so why did you add sin(t^2) at the end?

soroban used the product rule for derivatives to find v'(t) = a(t)
Click to expand...
 
I know that he used the product rule but if you start with t sin (t^2)

............ outside= t sin
............. inside= t^2

deriv= t cos (t^2)2t

do you add the original to the end?
 
the product of two functions is shown in red and blue ...

y = tsin(t<sup>2</sup>)

dy/dt = tcos(t<sup>2</sup>)*2t + sin(t<sup>2</sup>)*1

kapish?
 
fdragon said:
I know that he used the product rule but if you start with t sin (t^2)

............ outside= t sin
............. inside= t^2

deriv= t cos (t^2)2t

do you add the original to the end?
Click to expand...
You're trying to apply the chain rule to a product. sin(t^2) is derived using the chain rule; t*sin(t^2) is derived using the product rule, as Skeeter has shown.
 
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