Straight line formulae. Under constant acceleration v = Final velocity u= Initial velocity a = Acceleration t = Time. d = Distance v = u+ at d = ut + (1/2)at2 d = ((u + v)/2)t v 2 = u2 + 2ad Lets not worry how these formulae were derived. Lets simply get straight into using them with some examples. The steps you should follow for each example are: Step 1 - collect the data Step 2 - choose the appropriate formula Step 3 - give each pronumeral its respective value Step 3 - solve for the required variable. Exercise How long does it take for a car to cover 60m from a standing start, with a constant acceleration of 3.0m/s/s? Extra exercises of a similar nature to this one. Data u = 0m/s(Standing start) d = 60m a = 3.0m/s/s t =? Look at the data and choose a formula Formula and solution d = ut + (1/2)at2 60 =0 + 0.5 X 3 X t2 => 60/1.5 = t2 => square root of (40) = t => t = 6.32s A car can brake in an emergency with an acceleration of -9m/s/s. If the car is travelling at 27.7m/s in a straight line, what will be the distance it travels from the time the brakes are applied until it comes to rest? u = 27.7m/s d = ? a = -9.0m/s/s v = 0m/s (it stops) v 2 = u2 + 2ad 0 =(27.7)2 + 2 X -9 X d => -767.3 = -18d =>-767.3/-18 =42.63m => d = 42.63m A car is travelling at 4.0m/s due north, when the driver suddenly presses the accelerator to the floor. Ten seconds later he takes his foot off the accelerator and notices his speed is 40.0m/s. What is the acceleration of the vehicle during this period of time. u = 4.0m/s a = ? v = 40m/s t =10s v = u+ at 40.0 = 4.0 + a X 10 36.0 = 10a 3.6m/s/s = a A car travels along a straight road and passes a checkpoint at a speed of 23m/s. 23 seconds later it passes a second checkpoint travelling at 32m/s. How far apart are the two checkpoints? u = 23m/s a = ? v = 32m/s t =23s d = ((u + v)/2)t d = ((23 + 32)/2)23 d = 27.5 X 23 d = 632.5m Try this demonstration