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$v = -0.01t^3 + 0.22t^2 - 0.4t$. Find the 2 positive values of t for which the particle is instantaneously at rest. This part has been done: $t= 2$ and $t= 20$.

Find the distance travelled by the particle while its velocity is positive.

We find an expression for the distance $s$ and we just substitute $t=2$ and $t=20$ in it and then substract them together. My question is:

Why do we take these values of t?

ilove cupcakes
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1 Answers1

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Here's

Here is a plot of your given velocity function. Notice that the velocity is positive positive between $t=2$ and $t=20$, and so when you find the expression by integrating the velocity function, these two points of time become your limits of the integral. (Integral of the velocity is the displacement. In this case, since velocity is always positive, we are only travelling in one direction, so distance = displacement!)

Just to make this clear, $$s = \int_2^{20} v(t) dt = \int_2^{20} (-0.01t^3 + 0.22t^2 - 0.4t) dt = [f(t)]_{2}^{20} = f(20) - f(2)$$

where $f = -0.0025 t^4 + 0.0733333 t^3 - 0.2 t^2$

Arun Bharadwaj
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  • Thank you! but how could I have known that without plotting the graph? the question does not mention that you need to plot a graph and besides, does the graph become positive again before t= 0? – ilove cupcakes Feb 14 '21 at 19:48
  • oh, i get it you have to start from t =0. – ilove cupcakes Feb 14 '21 at 19:49
  • Here $t$ represents time, and since traversing through time backwards is not a thing (as of yet), we don't care about what happens before $t = 0$. For the other question about not graphing, solving for the time when the particle is at rest (part a) is going to give you the two time values where the particle is not moving. In between those two times however, the particle can never be at rest (can you think why?), so plugging in some point for $t$ between $2$ and $20$ should tell you if velocity is positive or negative (it will be positive, and so you'll know to integrate between 2 and 20) – Arun Bharadwaj Feb 14 '21 at 19:51