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An object's two dimensional velocity is given by #v(t) = ( 3t^2 - 5t , t )#. What is the object's rate and direction of acceleration at #t=2 #?

Answer 1

Nathan Ashcroft

#veca(t)=(7hati+hatj)m/s^2#

I'm guessing they're using normal convention replaced by brackets.

So, in vectorial form, #vec{v(t)}=(3t^2-5t)hati+thatj#

Acceleration is the time derivative of velocity, so you go that. #veca=d/dt(vecv(t))=d/dt(3t^2-5t)hati+d/dt(t)\hatj#

I'm sure you know differentiation, hence we'll get acceleration #veca(t)=(6t-5)hati+1hatj#

We need to find acceleration at time #t=2#, substitute, #veca(t)_{t=2}=(6*2-5)hati+hatj#

In the end, you'll get why the answer is as given in the box.

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Answer 2

Leo Abeyta

Rate of acceleration at (t=2): ( (6 , \text{m/s}^2, 2 , \text{m/s}^2) )

Direction of acceleration at (t=2): ( \theta = \arctan\left(\frac{2}{6}\right) )

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Answer from HIX Tutor

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