Proportion

y \propto \dfrac{1}{\sqrt{x}}, so we can write this as y = \dfrac{k}{\sqrt{x}}

 

We are given that y = 8, when x = 9, so substitute these into the equation and solve to find k:

 

\begin{aligned} 8 &= \dfrac{k}{\sqrt{9}} \\[1.2em] 8 &= \dfrac{k}{3} \\[1.2em] k &= 24 \end{aligned}

So, the equation is

y = \dfrac{24}{\sqrt{x}}

 

Then, find the value of x when y = 6, by substituting in y = 6 into the equation and solving for x:

 

\begin{aligned} 6 &= \dfrac{24}{\sqrt{x}} \\[1.2em] \sqrt{x} &= \dfrac{24}{6} \\[1.2em] \sqrt{x} &= 4 \\[1.2em] x &= 16 \end{aligned}

a) E \propto v^2, which can be written as E=kv^2

When v = 10, E = 3000, so

\begin{aligned} 3000 &= k \times 10^2 \\ 3000 &= 100k \\ k &= \dfrac{3000}{100} = 30 \end{aligned}

 

b) E = 30v^2

Substitute in v=4 into the equation to find the kinetic energy of the object if it is travelling at a velocity 4 \text{ m/s}:

 

E = 30 \times 4^2 = 30 \times 16 = 480 \text{ J}

a) W \propto d, which is equivalent to W = kd.

When d = 12, W = 60, therefore

60 = k \times 12 \Rightarrow k = 60 \div 12 = 5

 

b) W = 5d will be a straight line passing through the origin. You will only need to sketch the graph in the top right quadrant since distance cannot be negative.

(The gradient of the line is 5, but since we are only doing a sketch we do not need to write any values on the axes, so we can just draw any straight line passing through the origin with a positive gradient).

 

The graph will look like:

 

 

c) W = 5d

Substitute in d=25 into the equation to find the work done by the object if it is moved by 25 \text{ m}:

 

W = 5 \times 25 = 125 \text{ Nm}