HSC motion question (1 Viewer)

BlueGas · Jul 2, 2015

I need help with this question, I don't really know how to answer these sort of questions without equations.

InteGrand · Jul 2, 2015

BlueGas said:
I need help with this question, I don't really know how to answer these sort of questions without equations.

$(i) Velocity is given by the slope of the $x$-$t$ curve, since $v=\frac{\mathrm{d}x}{\mathrm{d}t}$. Hence $v=0$ precisely when the the $x$-$t$ curve has horizontal tangents, which we can see is roughly at $t=1$ and $t=5.5$.$

$(ii) Acceleration is given by the second derivative, i.e. $a=\frac{\mathrm{d}^2x}{\mathrm{d}t^2}$. Hence acceleration is zero when the curve has second derivative zero, which will be at the inflection point(s). This occurs around $t=3$, as we can see that here, the curve changes concavity.$

$(iii) The graph is plotting \emph{displacement} versus time, so \emph{distance} is at a maximum at the point where the function plotted has the greatest absolute value (i.e. furthest distance from the $x$-axis). We can see that this occurs around $t=5.5$ (same time as the second time that $v=0$, where the tangent is horizontal).$

$(iv) The velocity is given by the slope, so we are looking for when the slope is greatest. Note we are asked for maximum \emph{velocity}, not speed, so we want the greatest \emph{positive} slope. This occurs at $t=8$.$

BlueGas · Jul 2, 2015

InteGrand said:
$(i) Velocity is given by the slope of the $x$-$t$ curve, since $v=\frac{\mathrm{d}x}{\mathrm{d}t}$. Hence $v=0$ precisely when the the $x$-$t$ curve has horizontal tangents, which we can see is roughly at $t=1$ and $t=5.5$.$

$(ii) Acceleration is given by the second derivative, i.e. $a=\frac{\mathrm{d}^2x}{\mathrm{d}t^2}$. Hence acceleration is zero when the curve has second derivative zero, which will be at the inflection point(s). This occurs around $t=3$, as we can see that here, the curve changes concavity.$

$(iii) The graph is plotting \emph{displacement} versus time, so \emph{distance} is at a maximum at the point where the function plotted has the greatest absolute value (i.e. furthest distance from the $x$-axis). We can see that this occurs around $t=5.5$ (same time as the second time that $v=0$, where the tangent is horizontal).$

$(iv) The velocity is given by the slope, so we are looking for when the slope is greatest. Note we are asked for maximum \emph{velocity}, not speed, so we want the greatest \emph{positive} slope. This occurs at $t=8$.$

Thanks man, now I understand.

BlueGas · Jul 2, 2015

Need help with question c) ii)

leehuan · Jul 2, 2015

(ii) just uses an arithmetic/algebra trick.

$-2=2-4\\ x=\frac { t-2 }{ t+2 } \\ x=\frac { t+2 }{ t+2 } +\frac { -4 }{ t+2 } \\ x=1-\frac { 4 }{ t+2 } \\ \therefore x=1-4{ \left( t+2 \right) }^{ -1 }\\ v=4{ \left( t+2 \right) }^{ -2 }\\ v=\frac { 4 }{ { \left( t+2 \right) }^{ 2 } } \\ and\quad \ddot { x } =-\frac { 8 }{ { \left( t+2 \right) }^{ 3 } }$

BlueGas · Jul 2, 2015

leehuan said:
(ii) just uses an arithmetic/algebra trick.

$-2=2-4\\ x=\frac { t-2 }{ t+2 } \\ x=\frac { t+2 }{ t+2 } +\frac { -4 }{ t+2 } \\ x=1-\frac { 4 }{ t+2 } \\ \therefore x=1-4{ \left( t+2 \right) }^{ -1 }\\ v=4{ \left( t+2 \right) }^{ -2 }\\ v=\frac { 4 }{ { \left( t+2 \right) }^{ 2 } } \\ and\quad \ddot { x } =-\frac { 8 }{ { \left( t+2 \right) }^{ 3 } }$

Why did you do at the start -2 = 2 - 4

InteGrand · Jul 2, 2015

BlueGas said:
Why did you do at the start -2 = 2 - 4

That was to explain how he went from the second to the third line.

integral95 · Jul 2, 2015

Alternatively you could start with

$x = 1-\frac{4}{t+2} = \frac{t+2-4}{t+2} = \frac{t-2}{t+2}$

Flop21 · Jul 2, 2015

Hello, sorry to go on a tangent here:

but what actually is displacement? The x axis (vertical) in these graphs is displacement right? The horizontal axis, t, that's time right, so further along further the time. And the actual graph / curve is... "displacement versus time" what does this mean?

Sorry for the stupid questions. Just trying to get my head around this topic that we just started.

leehuan · Jul 3, 2015

By actual definition: Given a fixed point, the displacement of a particle from that fixed point is defined to be the shortest possible distance to reach that point, as well as the direction of the particle with respect to that point.

In the context of 2u maths, however.

Because the 2u course only demands you to know left and right, we define the displacement 'x' of a particle as how far it is to the left, or to the right of the particle. When x=0, we say that the particle is at the origin. When x>0, say x=1, we say the particle is 1m to the right of the origin. When x<0, say x=-3, we say the particle is 3m to the left of the origin. Note that our units for displacement is the metre.

Graphically, this means that at any point above the t-axis, the displacement of the particle is ? units to the right. Similarly for the left when it's below.

In our calculus-based course, we say that the particle's motion varies according to time. This means, that for every possible time we choose (t=0, t=1, t=244 etc.) the particle is at a certain displacement. That is why displacement is a function of time.
e.g. When x=sin(t) -> When t=0, x=0. So initially the particle is at the origin. When t=pi/2, x=1. So after pi/2 seconds have passed, the particle is 1m to the right of the origin. (Note that our unit for time is naturally seconds)

Displacement v.s. time just means displacement -> y-axis, and time -> x-axis
So instead of a y-axis and an x-axis, we have a new x-axis (representing displacement) and a t-axis (representing time).

EDIT: Occasionally they will use up and down instead of left and right. Although this is not common, if up and down are used just swap right and left with up and down respectively.

HSC motion question (1 Viewer)

BlueGas

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InteGrand

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BlueGas

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leehuan

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