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# Out in Space

Show that the acceleration ${dv\over dt}$ of a particle moving in a straight line can be written, in terms of its velocity $v$ and its displacement $x$ from a point of the line, in the form $v{dv\over dx}$.

At a distance $x\;\text{km}$ from the centre of the Earth the gravitational acceleration in $\text{km s}^{-2}$ is given by the formula $\frac{c}{x^2}$ where $c=4 \times 10^5$. If a space craft $10^4 \; \text{km}$ from the centre of the Earth is moving directly away from it at a speed of $10 \; \text{km s}^{-1}$ at what distance will it be moving with half that speed?

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### It's Only a Minus Sign

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Age 16 to 18

Challenge Level

Show that the acceleration ${dv\over dt}$ of a particle moving in a straight line can be written, in terms of its velocity $v$ and its displacement $x$ from a point of the line, in the form $v{dv\over dx}$.

At a distance $x\;\text{km}$ from the centre of the Earth the gravitational acceleration in $\text{km s}^{-2}$ is given by the formula $\frac{c}{x^2}$ where $c=4 \times 10^5$. If a space craft $10^4 \; \text{km}$ from the centre of the Earth is moving directly away from it at a speed of $10 \; \text{km s}^{-1}$ at what distance will it be moving with half that speed?

Solve these differential equations to see how a minus sign can change the answer