SUVAT equations question 10

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Chris Graham	said	Ready to use	8 years, 8 months ago
Amy Chadwick	said	Needs to be tested	8 years, 9 months ago

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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

Changed "distance m" t "distance in m" in part (b)

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Amy Chadwick 8 years, 9 months ago

Gave some feedback: Needs to be tested

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Amy Chadwick 8 years, 9 months ago

Created this.

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Question statement

At time $t=0$ a ball rolls through a point $A$ in the positive direction with a speed $\var{u} \mathrm{ms^{-1}}$ . The ball rolls with constant deceleration $\var{-a} \mathrm{ms^{-2}}$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

u	number	9
a	number	-2
t	number	9
s	number	20.25

Automatically regenerate variables?

Name

Data type

Value

A random number between and (inclusive) with step size

Description

initial velocity of the ball

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`

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Statement
Advice
"Part a)" - Minimum accepted value
"Part a)" - Maximum accepted value

Advice

First it is best to sketch an image of what is happening. The ball rolls through $A$ with constant deceleration and a positive velocity. As it is decelerating it will eventually stop (have velocity $0 \mathrm{ms^{-1}}$ ) at some unknown point, say $B$ . However the ball will continue to decelerate and therefore will change direction and return to the point $A$ .

a) We have $u=\var{u}$ and $a = \var{a}$ , where $a$ is negative as the ball is decelerating. We are asked to find the time, $t$ , that the ball returns to point $A$ , which means the balls displacement from $A$ will be zero. Therefore $s=0$ . We use the formula $s = ut+\frac{1}{2}at^2$ and solve for $t$ . As this is a quadratic in $t$ there will be two solutions, however we know that the ball is at point $A$ when $t=0$ so we are trying to find the other value of $t$ which is when the ball returns to point $A$ .

Therefore

$\begin{align} s &= ut+ \frac{1}{2}at^2, \\ \frac{1}{2}at^2 + ut &= 0, \\ t\left(\frac{1}{2}at+u\right) &= 0.\end{align}$

We discard the solution $t=0$ so

$\begin{align} \frac{1}{2}at + u &=0, \\ t & = -\frac{2u}{a}, \\ & = \var{precround(-(2*u)/a,3)} \mathrm{s}. \end{align}$
Therefore the ball returns to point

$A$ after

$\var{precround(-(2*u)/a,3)}$ seconds.

b) We have $u=\var{u}$ and $a=\var{a}$ . At the point labelled $B$ in the diagram the velocity of the ball is $v=0$ for an instant as it has reached its furthest point from $A$ and will change direction. We want to find the distance between point $A$ and $B$ and then double it, to calculate the total distance the ball travels as it goes from $A$ to $B$ and back. Therefore we use the equation $v^2 = u^2+2as$ , rearranged for $s$ .

$\begin{align} v^2 & = u^2 + 2as, \\ 2as & = v^2 - u^2, \\ s & = \frac{v^2 - u^2}{2a}, \\ & = \frac{ 0 - \var{u}^2}{2 \times \var{a}}, \\ & = \var{precround((-u^2)/(2*a),3)} \mathrm{m}. \end{align}$
So we have the distance from point

$A$ to point

$B$ ,

$s=\var{precround((-u^2)/(2*a),3)}$ metres. Therefore the total distance travelled by the ball in

$\var{precround(-(2*u)/a,3)}$ seconds is

$2 \times s=\var{precround(2*((-u^2)/(2*a)),3)}$ metres.

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Chris Graham 8 years, 8 months ago

Chris Graham commented 8 years, 8 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

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SUVAT equations question 10

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Chris Graham 8 years, 8 months ago

Chris Graham commented 8 years, 8 months ago

Amy Chadwick 8 years, 9 months ago

Amy Chadwick 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Number entry

Advanced settings

Generated value: `number`