Alan's copy of Numbas demo: Motion under gravity

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Alan Levine 5 years, 2 months ago

Published this.

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Alan Levine 5 years, 2 months ago

Created this as a copy of Numbas demo: Motion under gravity.

Name	Status	Author	Last Modified
Numbas demo: Motion under gravity	Ready to use	Christian Lawson-Perfect	05/10/2021 10:04
Liz's copy of Numbas demo: Motion under gravity	draft	Liz Thompson	03/02/2021 13:59
Ethan's copy of Numbas demo: Motion under gravity	draft	Ethan Smith	03/02/2021 13:59
Harry's copy of Numbas demo: Motion under gravity	draft	Harry Flynn	03/02/2021 13:59
Santa's elves equation of a straight line	Needs to be tested	Chris Graham	03/02/2021 13:59
Numbas demo: Motion under gravity	draft	Xiaodan Leng	03/02/2021 13:59
Alan's copy of Numbas demo: Motion under gravity	draft	Alan Levine	03/02/2021 13:59
Numbas demo: Images, videos and interactive diagrams	Ready to use	Christian Lawson-Perfect	24/08/2023 08:17
GEMIDDELD: Valbeweging van een balletje	draft	Alexander Holvoet	02/06/2024 14:36

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Question statement

A Numbas question can include interactive graphics, such as this plot of the trajectory given by the student's answer.

See this question in the public editor

A ball is thrown upwards, and moves according to the equation $\displaystyle{\frac{d^2z}{dt^2}=-g}$
(where $z(t)$ is distance in metres measured upwards from the ground and the constant acceleration of gravity, $g$ , is given as $9.81\;m/s^2$).

The ball is projected upwards with a speed $\var{v}\;m/s$.

Give any introductory information the student needs.

			Name	Type	Generated Value

t	number	5.8103975535
v	integer	57
mh	number	165.60
g	number	9.81
t1	number	5.81

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
V/g

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `number`

5.8103975535

← Depends on:

→ Used by:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
  Add
Part b) Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

{graphsolution()}

Input the vertical distance $z$ as a function of $t$.

Note that at $t=0$ we have $z=0$ and that $\displaystyle \frac{dz}{dt}=\var{v}m/s$.

Input gravitational acceleration as $g$.

$z=$ Unnamed gap

Your formula is plotted in the graph above. The vertical axis represents $z$ and the horizontal axis represents $t$.

Note that the blue line indicates that:

Your solution should go through $(0,0)$;
Your solution should have this line as the tangent to the curve at $(0,0)$, because $\displaystyle \frac{\mathrm{d}z}{\mathrm{d}t}=\var{v}\; m/s$.

Advice

a)

Integrating $\displaystyle{\frac{\mathrm{d}^2z}{\mathrm{d}t^2}=-g}$ once gives the velocity $\displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}=-gt+A}$.

But $A=\var{v}$ as the velocity is $\var{V}\;m/s$ at $t=0$.

So the velocity is

\begin{align} \frac{\mathrm{d}z}{\mathrm{d}t} &= \var{v}-gt & (1) \end{align}

Integrating again gives

\[ z = \var{v}t-\frac{g}{2}t^2+B \]

and $B=0$ as $z=0$ at $t=0$.

Hence the distance travelled upwards is given by

\begin{align} z &= \var{v}t-\frac{g}{2}t^2 & (2) \end{align}

{advicegraph()}

b)

The time $t_{\text{max}}$ taken to reach maximum height is the time satisfying $\displaystyle{\frac{dz}{dt}=0}$

$t_{\text{max}}$ is given from equation $(1)$ by

\begin{align}
\var{v} - gt_{\text{max}} &= 0 \\
gt_{\text{max}} &= \var{v} \\
t_{\text{max}} &= \frac{\var{v}}{g} \\[0.5em]
&= \frac{\var{v}}{9.81} \\[0.5em]
&= \var{t1}
\end{align}

(to $2$ decimal places)

This is at the point $A$ in the graph above.

c)

The maximum height $z_{\text{max}}$ is given from equation $(2)$ by substituting in the value $t_{\text{max}}= \var{v}/g$, giving

\begin{align}
z_{\text{max}} &= \var{v} \times \frac{\var{v}}{g} - \frac{g}{2}\left(\frac{\var{v}}{g}\right)^2 \\
&= \frac{\var{v}^2}{g}-\frac{g\var{v}^2}{2g^2} \\
&= \frac{\var{v}^2}{2g} \\
&= \var{mh}\;m
\end{align}

(to $2$ decimal places)

This is at the point $B$ in the graph above.

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Mathematical expression
Mathematical expression		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet

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Alan's copy of Numbas demo: Motion under gravity

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Alan Levine 5 years, 2 months ago

Alan Levine 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: `number`

Parts

Gap-fill

a)

b)

c)

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Error

Alan's copy of Numbas demo: Motion under gravity

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Alan Levine 5 years, 2 months ago

Alan Levine 5 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

a)

b)

c)

Generated value: `number`