Figure 5 is a speed-time graph of an object. Given that the total distance travelled by the object is 7200m, calculate the acceleration in the first 40 seconds.

The total distance travelled is the area under the speed-time graph. The graph forms a trapezium.
The parallel sides of the trapezium are the time intervals where the speed is non-zero.
Length of the first parallel side (top) = $100\, \text{s} - 40\, \text{s} = 60\, \text{s}$.
Length of the second parallel side (bottom) = $120\, \text{s} - 0\, \text{s} = 120\, \text{s}$.
The height of the trapezium is the maximum speed, $V$.
The formula for the area of a trapezium is $\frac{1}{2} (a+b)h$.

$$\text{Distance} = \frac{1}{2} (60\, \text{s} + 120\, \text{s}) V$$ $$\text{Distance} = \frac{1}{2} (180\, \text{s}) V$$ $$\text{Distance} = 90V$$

We are given that the total distance travelled is $7200\, \text{m}$.

$$90V = 7200\, \text{m}$$ $$V = \frac{7200}{90}\, \text{m/s}$$ $$V = 80\, \text{m/s}$$

Acceleration is the gradient of the speed-time graph. In the first 40 seconds, the object accelerates from rest ($u = 0\, \text{m/s}$) to a final speed of $V = 80\, \text{m/s}$ over a time interval of $t = 40\, \text{s}$.
The formula for acceleration is $a = \frac{v - u}{t}$.

$$a = \frac{80\, \text{m/s} - 0\, \text{m/s}}{40\, \text{s}}$$ $$a = \frac{80\, \text{m/s}}{40\, \text{s}}$$ $$a = 2\, \text{m/s}^2$$

The acceleration in the first 40 seconds is $\boxed{\text{2 m/s}^2}$.

19. Make $u$ the subject of the formula $A = \frac{1}{2} m (v^2 - u^2)$

Add $u^2$ to both sides and subtract $\frac{2A}{m}$ from both sides.

$$u^2 = v^2 - \frac{2A}{m}$$

$$u = \pm \sqrt{v^2 - \frac{2A}{m}}$$

The subject of the formula is $\boxed{u = \pm \sqrt{v^2 - \frac{2A}{m}}}$.