Particle displacement

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From users who are members of MfEP Progress Quizzes :

Hanh Vo

said

Ready to use

3 years, 4 months ago

History

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Don Shearman 11 months, 2 weeks ago

Published this.

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Hanh Vo 3 years, 4 months ago

Gave some feedback: Ready to use

○

Don Shearman 3 years, 8 months ago

Created this as a copy of Ball height.

Name	Status	Author	Last Modified
Ball height	Ready to use	Don Shearman	16/04/2024 03:06
Particle displacement	Ready to use	Don Shearman	08/05/2024 05:57
Bending moments	Ready to use	Don Shearman	16/04/2024 03:06

There are 3 other versions that do you not have access to.

Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

a	number	1.9
b	number	5.7
c	integer	75
ans	number	3.4645414676
ans2	number	-3.797874801

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(0.5..1.9 #0.1 except 1)

JME function reference

Description

coefficient of t in quadratic - represents initial velocity of particle

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`

1.9

→ Used by:

ans
ans2

Advice
"Unnamed part" - prompt
"Unnamed part" → "Unnamed gap" - Minimum accepted value
"Unnamed part" → "Unnamed gap" - Maximum accepted value

Parts

Gap-fill
1. Gap Number entry
  Add

Gap-fill

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

The displacement, $s$ , of a particle (in metres) is given by

$\simplify{s={a}t+{b}t^2}\quad t\ge 0$ where

$t$ is time in seconds.

Find the time taken for the particle to have a displacement of $\var{c}$ m. Write your final answer in the box below and show full working on your handwritten notes.

$t=$ Unnamed gapsec (round answer to 2 decimal places)

Advice

We require $s=\var{c}$ , hence we need to solve the equation

$\simplify{{b}t^2+{a}t}=\var{c}$ or equivalently

$\simplify{{b}t^2+{a}t-{c}}=0$

Using the quadratic formula we have

$\begin{align} t&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-\var{a}\pm\sqrt{\simplify[!collectNumbers]{{a}^2-4*{b}*{-c}}}}{\var{2*b}}\\&=\frac{-\var{a}+\sqrt{\simplify{{a}^2-4*{b}*{-c}}}}{\var{2*b}}\text{ or }\frac{-\var{a}-\sqrt{\simplify{{a}^2-4*{b}*{-c}}}}{\var{2*b}}\\&=\var{ans}\text{ or }\var{ans2}\end{align}$

Since we are told that $t\ge 0$ we can ignore the negative answer and conclude that particle will have a displacement of $\var{c}$ m when $t$ is approximately $\var{precround({ans},2)}$ seconds.

Give a worked solution to the whole question.

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

This question is used in the following exams:

SCC Class Test 1 by Hanh Vo in SCC PrelimMaths.
Don's copy of Class Test 1 for testing by Don Shearman in Don's workspace.

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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Checkpoint description

Don Shearman 11 months, 2 weeks ago

Hanh Vo 3 years, 4 months ago

Don Shearman 3 years, 8 months ago

Error in variable testing condition

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Warning

Generated value: `number`

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Particle displacement

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Don Shearman 11 months, 2 weeks ago

Hanh Vo 3 years, 4 months ago

Don Shearman 3 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Generated value: `number`