Straight-Line Graphs: Question 4

Worked solution

Part (a): Find the gradient

The gradient $m$ of a straight line through two points $(t_1, V_1)$ and $(t_2, V_2)$ is the change in $V$ divided by the change in $t$:

$m = \frac{V_2 - V_1}{t_2 - t_1}$

Take $(2,\,630)$ as $(t_1, V_1)$ and $(6,\,450)$ as $(t_2, V_2)$:

$m = \frac{450 - 630}{6 - 2} = \frac{-180}{4} = -45$

$\boxed{m = -45}$

The gradient is negative because the volume is decreasing as time increases: the tank is emptying.

Tip: It does not matter which point you label “1” and which “2”, provided you are consistent top and bottom. Reversing both gives $\dfrac{630 - 450}{2 - 6} = \dfrac{180}{-4} = -45$, the same answer.

Part (b): Equation in the form $V = mt + c$

We have $m = -45$, so the line is:

$V = -45t + c$

To find $c$, substitute the coordinates of either point (the line passes through both). Using $(2,\,630)$:

$630 = -45(2) + c$ $630 = -90 + c$ $c = 630 + 90 = 720$

So the equation is:

$\boxed{V = -45t + 720}$

Check with the other point $(6,\,450)$:

$-45(6) + 720 = -270 + 720 = 450 \checkmark$

This matches, confirming the equation is correct.

Part (c): Interpret the intercept and the gradient

Starting volume. The outlet was opened at $t = 0$. Substituting $t = 0$ into the equation gives $V = c$, so the intercept is the starting volume:

$V = -45(0) + 720 = 720$

The tank held $\boxed{720\text{ litres}}$ when the outlet was opened.

Meaning of the gradient. The gradient measures how $V$ changes for each extra minute. Here $m = -45$, and because $V$ is in litres and $t$ is in minutes, this means:

the volume is decreasing by $45$ litres every minute (the tank drains at a steady rate of $45$ litres per minute).

Part (d): The second tank

The second tank drains at the same rate, so its line has the same gradient $m = -45$ (the two lines are parallel). It starts full with $900$ litres at $t = 0$, so its intercept is $c = 900$:

$\boxed{V = -45t + 900}$

The tank is empty when $V = 0$. Set $V = 0$ and solve for $t$:

$0 = -45t + 900$ $45t = 900$ $t = \frac{900}{45} = 20$

$\boxed{t = 20\text{ minutes}}$

Check: at $t = 20$, $\; V = -45(20) + 900 = -900 + 900 = 0 \checkmark$. The second tank is empty after $20$ minutes.