Motion - Straight line formulae

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)at^{2 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

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