KINEMATICS. Definitions average velocity = vav = average acceleration = aav = Straight line graphs Graph type Found from.
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1 KINEMATICS Definitions average velocity = totaldisplacement total time vav = x x1 Straight line graphs Graph type Found from Direct reading Gradient average acceleration = 'x' at any 't' 't' at any 'x' change inveloci ty time taken aav = v - u x - t v - t a - t Instantaneous velocity at any point. Vav between any two points 'v' at any 't' 't' at any 'v' Instantaneous 'a' Average 'a' Area under graph Meaningless x v 'a' at any 't' 't' at any 'a' Meaningless When given a graph in the exam, look at three things on the graph before even reading the question: type of graph (x - t, v - t, etc.) the units on the axis the limit reading on each axis. The following relationships hold for straight line motion. v - u v a = = v = u at v = u ax x = ut t 1 at (u v) x = t t = 0 is the beginning of the time interval being considered, i.e. the instant at which 'u' occurs. that when using the formulae: 3 numerical facts are needed a negative answer for 't' indicates a time previous to 't' = 0. x is not necessarily the same measure as the total distance travelled a body that is travelling in one direction and accelerating in the opposite direction is slowing down. when given the distance travelled in a certain time interval, this distance is the instantaneous velocity halfway through the time interval. Eg. if a body travels 14 m in the seventh second ('t' = 6 to 't' = 7 sec) then the actual velocity at 6.5 seconds is 14 m/s. for motion along the horizontal it is usual to take 'to the right positive' for vector sense for vertical motion (bodies projected vertically or dropped from rest) the direction of the initial displacement is usually taken as positive for vertical motion, the acceleration (symbolised by 'g') is 10 m/s vertically downwards at all times, even if the body is momentarily at the top of its vertical flight. Year 1 Physics Page 1
2 Graphs for constant acceleration Displacement Velocity Acceleration v g u time Time time t Change in velocity v = vf - vi, remember that these are vector v quantities, so direction is vital. The acceleration, a =, will always be in the direction of v. Year 1 Physics Page
3 VCAA Exam Examples 00 In a road test, a car was uniformly accelerated from rest over a distance of 400 m in 19.0 s. The driver then applied the brakes, stopping the car in 5.1 s with constant deceleration. Question 1 Calculate the acceleration of the car for the first 400 m.. Q1 Distance travelled x = ut + 1 at x Given u = 0, x = 400 and t = 19.0 a = t = =. ms-. This question was an example of uniformly accelerated motion and application of the equation s = ut + ½at resulted in an answer of. m s -. The most common error, committed by a large number of students, was to start by calculating the average speed via 400 = 1.05 m s-1 19 and then calculate the acceleration as 1.05 = 1.1 ms - 19 Question Calculate the average speed of the car for the entire journey, covering both the acceleration and braking sections.. Average speed = total distance/total time. I would use a velocity-time graph sketch v area = t Left hand triangle has an area of 400 [displacement] Right hand triangle has same height, but base of 5.1 instead of 19 [stopped in 5.1, accelerated for 19] Thus area is (400 x 5.1)/19 Total area = total distance = (4.1 x 400)/19 Average speed = distance / time = [(4.1 x 400)/19]/4.1 = 400/19 = 1.1 ms -1 This question could be solved in a number of ways. The most common, and longest, method was to find the distance and time for each of the accelerating and braking phases and then divide the total distance by the total time for the average speed. A simpler method was to realise that the average speed for both sections (accelerating and braking) was the same and hence all one had to do was find the final speed for the accelerating section (4.1 ms -1 ) and then halve it. giving an answer of 1 m s -1 for the average speed. Year 1 Physics Page 3
4 Very few students recognised the symmetry of the uniform acceleration and braking sections and so took longer than anticipated in solving this question. A number of students chose 1.1 m s -1 as the maximum speed rather than 4. m s -1. This reinforced the fact that many students are unable to distinguish between average and instantaneous velocity. Another common error was to assume that the braking time was the same as the accelerating time of 19 s. The graphs (A F) in the key below should be used when answering Questions 3 and 4. The horizontal axis represents time and the vertical axis could be velocity or distance. Question 3 Which of the graphs (A F) best represents the velocity time graph of the car for the entire journey?. Best graph is B because it matches the one I used. Also it is the only one which has two fixed gradients, corresponding to the two sections of constant acceleration Graph B best represented the velocity-time graph for the car for the entire journey. It showed a uniform increase in speed when accelerating and a uniform decrease in speed when braking. The most common incorrect response was to choose graph D. Question 4 Year 1 Physics Page 4
5 Which of the graphs (A F) best represents the distance time graph of the car for the entire journey? Best graph is E because it needs to always increasing, since the car is moving over a distance of 400m. Also it is the only one that is always curved, corresponding to the two sections of constant acceleration Graph E best represented the distance-time graph for the car for the entire journey. There was a disappointing rate of correct responses for such a straightforward question. The most common incorrect response was to choose graph C or F. 001 A recent car advertisement states: you can legally floor the throttle for 5.9 seconds. This means that it takes 5.9 s to reach a speed of 100 km h -1 starting from rest. Assume that this situation can be modelled as uniform acceleration. Question 1 Calculate the magnitude of the uniform acceleration of this car and state the unit of this acceleration. To convert from km/hr to m/s you need to divide by 3.6. (This should be on your cheat sheet) v km/hr = 7.8 m/s. v = 7.8 m/s. acc = = = 4.71 ms Application of the equations for uniform acceleration resulted in an answer of either 4.7 m s - or 16.9 km h -1 s -1. This question was done quite well with most students obtaining the correct magnitude, but with less giving the correct unit for acceleration. The most common error was in giving the magnitude answer of 16.9 with the corresponding incorrect unit of m s -. Question 13 Calculate the distance travelled in this time of 5.9 seconds Distance travelled x = ut + 1 at 1 x = = 81.9 m = 8m Application of the equations for uniform acceleration, using the acceleration value of 4.7 m s -, resulted in an answer of 8 m. This question was reasonably well done with most students using the correct equation for uniform acceleration to calculate the distance travelled. The most common error arose as a result of mixing up the magnitude and unit for acceleration. Year 1 Physics Page 5
6 000 The standing 400 m time for a car is the time that it takes to travel 400 m on a level road, accelerating from rest. The standing 400 m time of a car was 16.0 s. Question 1 Calculate the acceleration of the car, assuming constant acceleration for the entire journey x ut at a 16 a a 3.1ms 16 The acceleration of the car was calculated from the equations for uniform acceleration to be 3.1 m s -. The most common error here was first to calculate the average velocity and then use this value in one of the equations for uniform acceleration. This was a remarkably common error. Question Assuming constant acceleration, calculate the speed of the car at the end of the 400 m. 1 1 x vt at 400 v v 800 v 50ms 1 16 The speed of the car was calculated via the equations for uniform acceleration, resulting in an answer of 50 m s -1. There were a number of correct consequential answers for students who obtained the incorrect answer to Question 1. This question was usually correctly answered, either directly or consequentially. Year 1 Physics Page 6
7 1998 The figure below appeared in a newspaper featuring skydiving from an aircraft. In this particular example the total mass of the skydiver and equipment is 100 kg. The skydiver jumps from a height of 3000 m above the ground and reaches a constant terminal velocity of 190 km h -1 in a time of 15 s. She then falls at this constant speed of 190 km h -1 for a further 35 s before opening the parachute. Question 1 Convert 190 km h -1 into m s m/s To convert from km/hr to m/s you need to divide by 3.6. (This should be on your cheat sheet) = 5.8 m/s A simple question to start off the exam. A speed of 190 km h -1 converts to 53 m s -1. Seventy-four per cent of students were able to correctly convert this speed, with the most common errors being a failure to convert kilometres to metres, or to use the reciprocal of the correct conversion factor. Year 1 Physics Page 7
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