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speed (m s - 1 ) 25 20 15 10 5 0 QUESTION TWO A family drives their 800 kg car to a holiday house to enjoy their summer holiday. Part of their journey is represented by the speed-time graph shown below: (a) Calculate the acceleration acting during the first 20 seconds. (b) Section A 0 20 40 60 80 100 120 Calculate the Work done during the first 20 seconds time (s) QUESTION TWO ANSWERS Part a) and b) Achievement is correct calculation of acceleration OR calculation of Force (using incorrect value for a) OR correct calculation of distance travelled. Units must be given. a = gradient of line = change in v / t = 25 / 20 = 1.25 m s -2 Merit is correct calculation of acceleration AND correct calculation of Force (using incorrect value for a) OR correct calculation of distance travelled. Units must be given. W = Fd F = ma = 800 x 1.25 = 1000 N d = area under line for first 20 seconds = ½ x base x height d = 0.5 x 20 x 25 = 250 m Excellence is correct answer for Work with all working shown and units W = Fd = 1000 x 250 = 250 000 J Part C Achievement would be to give two or three of the points given below: The car has a constant velocity (is not accelerating) The total force acting on the car is zero The forces on the car are balanced Work = Change in energy E k = ½ mv 2 = ½ x 800 x 25 2 = 250 000 J (c) Explain (in a <strong>Physics</strong> sense) why the graph shows that there was no Work done between time = 20 seconds and time = 60 seconds. Include in your answer: The motion of the car during that time and what that tells us about the forces involved What work is At least one energy calculation Merit would be to give at least one of the points listed below Because the car has a constant velocity it is not accelerating (a = 0) and so the total Force acting on the car is 0 (F = ma = m x 0 = 0) Work = Fd and as F = 0 then Work = 0 x distance = 0 J Work = Change in energy and the energy is not changing between t = 20 s and t = 60 s because the car has the same (Kinetic) energy over that time as velocity is not changing (E k = ½ mv 2 ) Excellence would be to give at least two of the points required for Merit
QUESTION THREE While a 1200 kg car is moving at 20 m s -1 , a dog runs across the road in front of it. When the driver applies the brakes, the car stops in a distance of 25 metres. a) Calculate the average force acting on the car when the driver applies the brakes. Your answer should include: Any energy transformations that take place Any assumptions you have made All calculations with working an appropriate unit with your answer. b) Use your answer from a) to calculate the time it takes for the car to stop c) If the car had been travelling uphill would it have taken a greater or lesser distance to stop? Discuss your answer in terms of: The forces acting on the car Any energy transformations that take place QUESTION THREE ANSWERS Achievement would be to give two of the points given below: Kinetic energy is changed into heat energy (Work done by friction) Assuming all Kinetic energy is changed into Heat (Work done by friction) Work done to stop car = Kinetic energy of car E k = ½ mv 2 = ½ x 1200 x 20 2 = 240 000 J (Working must be shown) (Or Merit calculation done correctly with incorrect E k value) Merit is all points for Achieved as well as force calculation done W = Fd = 240 000 so F = 240 000 / d = 240 000 / 25 = 9600 N Part b) Achievement is one calculation done correctly F = ma so a = F / m = 9600 / 1200 = 8 m s -2 (Or carried error using your answer from a)) a = change in v / t so t = change in v / a = 20 m s -1 / 8 = 2.5 s (Or carried error if the calculation of a is incorrect) Merit is both calculations done correctly Part c) Achievement would be to give two of the points given below: The car would take less distance to stop The forces slowing the car would be bigger so the car would stop quicker (in less distance) OR The weight force of the car is partly acting backwards slowing the car Kinetic energy is being turned into heat and sound AND Gravitational Potential Energy Merit would be to give at least one of the points listed below Because the weight force is (partly) acting backwards this means that the total force stopping the car will be bigger than just the force being applied by the brakes On the flat all the E k of the car would have to be turned into heat and sound by the brakes. But as the car is going uphill some of the E k is being turned into E p so there is less to be transformed by the brakes so the car stops sooner. Excellence is to extend the points given for Merit If F (back) is bigger; than a = F/m is bigger and v will decrease to 0 faster. If some E k is turned into Ep then W (done by brakes) will be less and since w = Fd then d = W / F so smaller W = smaller stopping distance
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