(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) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph (i) acceleration; (ii) retardation; (iii) total distance covered.
(b) Show that the displacement of a body moving with uniform acceleration a is given by \(s = ut + \frac{1}{2} at^{2}\) where u is the velocity of the body at time t = 0.
(c) A particle moving in a straight line with uniform deceleration has a velocity of 40ms\(^{-1}\) at a point P, 20ms\(^{-1}\) at a point Q and comes to rest at a point R where QR = 50m. Calculate the: (i) distance PQ; (ii) time taken to cover PQ; (iii) time taken to cover PR.
(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.
Velocity-time graph: gradient of AB gives acceleration, gradient of CD gives retardation, and the shaded area O-A-B-C-D gives the total distance covered.
(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 \(v^{2} = u^{2} - 2as\):
\[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]
(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.
Velocity-time graph: gradient of AB gives acceleration, gradient of CD gives retardation, and the shaded area O-A-B-C-D gives the total distance covered.
(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 \(v^{2} = u^{2} - 2as\):
\[ 0 = 20^{2} - 2a(50) \;\Rightarrow\; 400 = 100a \;\Rightarrow\; a = 4\,\text{ms}^{-2}\ (\text{deceleration}). \]
