(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.
Explanation
(a)
Velocity- time graph:
(i) Acceleration : Gradient of AB in the graph above; (ii) retardation: Gradient of CD in the graph above; (iii) total distance covered is the area under the graph.
(b) S = average velocity x time
(S = (frac{u +v}{2}) times t … (1))
but (v = u + at … (2)), substitute (2) in (1)
(S = (frac{u + u + at}{2}) times t)
(S = (frac{2u + at}{2}) times t )
(S = ut + frac{1}{2} at^{2})
(c) P ———– Q —-50m——- R
40 ms(^{-1}) 20 ms(^{-1})
Since the acceleration is uniform and using equation
(v^{2} = u^{2} + 2as) for the motion along QR.
(0 = 20^{2} – 2a(50)) (retardation)
(0 = 400 – 100a implies 100a = 400)
(a = 4 ms^{-2}) (retardation, hence it is negative)
(20^{2} = 40^{2} + 2(-4) s)
(400 = 1600 – 8s implies 8s = 1200)
(s = 150m)