Home > Calculus > Application of the Second Derivative (Acceleration)

A particle moves according to the equation #s=1-1/t^2#, how do you find its acceleration?

Answer 1

Assuming #s# is the position function, recall that acceleration is the second derivative of position. Therefore, take two derivatives in a row to obtain the acceleration.

The first derivative gives you velocity, and the second gives you acceleration.

#(ds)/(dt) = vecv = 2/t^3#

#color(blue)(veca) = (d^2s)/(dt^2) = d/(dt)[(ds)/(dt)] = d/(dt)[2/t^3] = color(blue)(-6/t^4)#

What acceleration does the particle have after #"3 s"# have passed? Is it slowing down or speeding up?

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Answer 2

Charles Arocha

To find the acceleration, you need to differentiate the equation for position (s) twice with respect to time (t) to get the equation for acceleration (a).

Given ( s = 1 - \frac{1}{t^2} ),

First, find the velocity (v) by differentiating ( s ) with respect to ( t ):

[ v = \frac{ds}{dt} = \frac{d}{dt} \left(1 - \frac{1}{t^2}\right) ]

Then, differentiate ( v ) with respect to ( t ) to find acceleration (a):

[ a = \frac{dv}{dt} = \frac{d^2s}{dt^2} = \frac{d^2}{dt^2} \left(1 - \frac{1}{t^2}\right) ]

Now, compute ( a ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.