Could someone check this because I have different answers to those given for parts ii and iii.

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- Dec 8th 2012, 02:44 AMStuck Manv-t graph
Could someone check this because I have different answers to those given for parts ii and iii.

- Dec 8th 2012, 03:09 AMMarkFLRe: v-t graph
i) You want to use $\displaystyle a=\frac{\Delta v}{\Delta t}$. Your result will be in $\displaystyle \frac{\text{km}}{\text{min}\cdot\text{hr}}$. To make the required conversion to the units given use:

$\displaystyle x\frac{\text{km}}{\text{min}\cdot\text{hr}}\cdot \frac{1000\text{ m}}{1\text{ km}}\cdot\frac{1\text{ hr}}{3600\text{s}}\cdot\frac{1\text{ min}}{60\text{ s}}=\frac{x}{216}\,\frac{\text{m}}{\text{s}^2}$

ii) The area under the velocity function is two right triangles, a trapezoid and a rectangle. Then divide the area by 60 to convert the minutes to hours to get the result in km.

iii) Use $\displaystyle \bar{v}=\frac{d}{t}$ - Dec 8th 2012, 03:41 AMStuck ManRe: v-t graph
The unit conversion has been tricky for me. Its not x/216 its 1000x/60^3. Your left hand side seems to be ok. I had not the trapezoid (trapezium to non americans).

- Dec 8th 2012, 03:55 AMStuck ManRe: v-t graph
x/216 is equal to 1000x/60^3.