(a) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance ...

(a) Obtaining quantities from a velocity-time graph

On a velocity-time graph the velocity is plotted on the vertical axis and time on the horizontal axis. A typical graph rises from A to B, stays level from B to C, then falls from C to D as shown below.

(i) Acceleration is obtained from the gradient of the rising portion AB. If the velocity increases from \(v_1\) to \(v_2\) as time changes from \(t_1\) to \(t_2\) along AB, then \[ a = \text{gradient of AB} = \frac{v_2 - v_1}{t_2 - t_1} \quad (\text{a positive slope}). \]

(ii) Retardation (deceleration) is obtained from the gradient of the falling portion CD. Here the velocity decreases with time, so the slope is negative: \[ \text{retardation} = -\,(\text{gradient of CD}) = \frac{v_C - v_D}{t_D - t_C}. \]

(iii) Total distance covered is the total area enclosed between the graph line and the time axis (the shaded region O-A-B-C-D). It is found by splitting the region into triangles and a rectangle (or one trapezium) and adding the areas.

(b) Deriving \(s = ut + \tfrac{1}{2}at^{2}\)

The displacement equals the area under the velocity-time graph, which for uniform acceleration is a trapezium of parallel sides \(u\) (initial velocity) and \(v\) (final velocity) and width \(t\): \[ s = \text{average velocity} \times t = \left(\frac{u + v}{2}\right)t. \tag{1} \] For uniform acceleration \(a\), the final velocity is \[ v = u + at. \tag{2} \] Substituting (2) into (1): \[ s = \left(\frac{u + (u + at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t \] \[ \boxed{\,s = ut + \tfrac{1}{2}at^{2}\,}. \]

(c) Uniform deceleration problem

Along the straight line the velocities are: at P, \(u = 40\,\text{ms}^{-1}\); at Q, \(20\,\text{ms}^{-1}\); at R the particle is at rest (\(0\,\text{ms}^{-1}\)); and \(QR = 50\,\text{m}\).

Find the deceleration using the motion from Q to R, with \(v^{2} = u^{2} - 2as\): \[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]

(i) Distance PQ (from P to Q): \[ 20^{2} = 40^{2} - 2(4)(PQ) \;\Rightarrow\; 400 = 1600 - 8\,PQ \;\Rightarrow\; PQ = \frac{1200}{8} = 150\,\text{m}. \]

(ii) Time taken to cover PQ, using \(v = u - at\) from P to Q: \[ 20 = 40 - 4t \;\Rightarrow\; 4t = 20 \;\Rightarrow\; t = 5\,\text{s}. \]

(iii) Time taken to cover PR (from P to rest at R): \[ 0 = 40 - 4t \;\Rightarrow\; 4t = 40 \;\Rightarrow\; t = 10\,\text{s}. \]