SUVAT equations question 3

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From users who are members of Maths Support Wiki - Mechanics :

Chris Graham	said	Ready to use	8 years, 8 months ago
Amy Chadwick	said	Needs to be tested	8 years, 10 months ago

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Chris Graham commented 4 years, 1 month ago

Added an additional restriction to ensure that part d) result is always positive

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Chris Graham 8 years, 8 months ago

Gave some feedback: Ready to use

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Chris Graham commented 8 years, 8 months ago

Small changes to advice to make decimal/fraction answers more consistent.

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Amy Chadwick 8 years, 10 months ago

Gave some feedback: Needs to be tested

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Amy Chadwick 8 years, 10 months ago

Gave some feedback: Has some problems

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Amy Chadwick 8 years, 10 months ago

Created this.

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Question statement

A cyclist rides in a straight line with constant decelaration $\var{-{a}} \mathrm{ms^{-2}}$ from a point $A$ to a point $B$ . At point $A$ the cyclist's speed is $\var{u}\mathrm{ms^{-1}}$ and at point $B$ the speed is $\var{v}\mathrm{ms^{-1}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

u	integer	8
a	number	-0.75
v	integer	3
t	number	6.6666666667
t2	integer	10
v2	number	0.5
s	number	42.5

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

initial speed

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

Statement
Advice
Variable testing condition
"Part a)" - Minimum accepted value
"Part a)" - Maximum accepted value
"Part b)" - Minimum accepted value
"Part b)" - Maximum accepted value
"Part c)" - Minimum accepted value
"Part c)" - Maximum accepted value
"Part d)" - Minimum accepted value
"Part d)" - Maximum accepted value

Advice

a) You are given $u = \var{u},$ $v=\var{v}$ , $a = \var{a}$ , $t = ?$ , $s = ?$

Note that $a$ is negative because the cyclist is decelerating.

You need $t$ , so you use $v=u+at$ , rearranged for $t$ .

$\begin{align} t & = \frac{v-u}{a}, \\ & = \frac{\var{v}-\var{u}}{\var{a}}, \\ & = \var[fractionNumbers]{(v-u)/a} \mathrm{s}. \end{align}$

The time taken to move from $A$ to $B$ is $\var[fractionNumbers]{(v-u)/a}$ seconds.

b) You need the distance, $s \mathrm{m}$ . Therefore you can use $s = \left(\frac{u+v}{2}\right)t$ , with the $t$ value found in part a).

$\begin{align} s & = \left(\frac{u+v}{2}\right)t, \\ & = \left(\frac{\var{u}+\var{v}}{2}\right) \times \var[fractionNumbers]{(v-u)/a}, \\ & = \var[fractionNumbers]{(u+v)*0.5*((v-u)/a)} \mathrm{m}. \end{align}$
The distance from

$A$ to

$B$ is

$\var[fractionNumbers]{(u+v)*0.5*((v-u)/a)}$ metres.

c) You now have $u=\var{u}$ , $a = \var{a}$ , $t=\var{t2}$ , $s=?$ and you are trying to find the velocity at point $C$ so $v=?$

Therefore you can use $v = u+at$ .

$\begin{align} v & = u + at, \\ & = \var{u} + \left(\var{a} \times \var{t2}\right), \\ & = \var[fractionNumbers]{u+a*t2} \mathrm{ms^{-1}}. \end{align}$

Note that the velocity can be negative - this means the cyclist is moving from right to left. Therefore the velocity of the cyclist at point $C$ is $\var{-(u+a*t2)} \mathrm{ms^{-1}}$ in the direction $\simplify{vec:BA}.$

d) We now need the distance, $s \mathrm{m}$ . We have $u=\var{u}$ , $t=\var{t2}$ and we have just evaluated the final velocity at point C, $v=\var{u+a*t2}$ .

$\begin{align} s & = \left(\frac{u + v}{2}\right)t, \\ & = \left(\frac{\var{u} + \var{u+a*t2}}{2}\right) \times \var{t2}, \\ & = \var[fractionNumbers]{(u+u+a*t2)*0.5*t2} \mathrm{m}. \end{align}$

The distance from A to C is $\var[fractionNumbers]{(u+u+a*t2)*0.5*t2}$ metres.

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Chris Graham commented 4 years, 1 month ago

Chris Graham 8 years, 8 months ago

Chris Graham commented 8 years, 8 months ago

Amy Chadwick 8 years, 10 months ago

Amy Chadwick 8 years, 10 months ago

Amy Chadwick 8 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Number entry

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SUVAT equations question 3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Chris Graham commented 4 years, 1 month ago

Chris Graham 8 years, 8 months ago

Chris Graham commented 8 years, 8 months ago

Amy Chadwick 8 years, 10 months ago

Amy Chadwick 8 years, 10 months ago

Amy Chadwick 8 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Number entry

Advanced settings

Generated value: `integer`