神马作文网1.A cyclist comes to the top of a hill 165 metres long trave
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1.A cyclist comes to the top of a hill 165 metres long travelling at 5 ms(-1次方),and free-wheels down it with an acceleration of 0.8ms(-2次方).Write expressions for his speed and the distance he has travelled after t seconds.Hence find how long he takes to reach the bottom of the hill,and how fast he is then travelling.
2.A freight train 1/4 KM long takes 20 seconds to pass a signal.The train is decelerating at a constant rate,and by the time the rear truck has passed the signal it is moving 10 kilometres per hour slower than it was when the front of the train passed the signal.Find the deceleration in kilometre-hour units,and the speed at which the train is moving when the rear truck has just passed the signal.
3.A cyclist is free-wheeling down a long straight hill.The times between passing successive kilometre posts are 100 seconds and 80 seconds.Assuming his acceleration is constant,find this acceleration.
2.A freight train 1/4 KM long takes 20 seconds to pass a signal.The train is decelerating at a constant rate,and by the time the rear truck has passed the signal it is moving 10 kilometres per hour slower than it was when the front of the train passed the signal.Find the deceleration in kilometre-hour units,and the speed at which the train is moving when the rear truck has just passed the signal.
3.A cyclist is free-wheeling down a long straight hill.The times between passing successive kilometre posts are 100 seconds and 80 seconds.Assuming his acceleration is constant,find this acceleration.
1. Let v represent the cyclist's speed, and d the distance the cyclist has travelled. Then v = 5 + 0.8t, and d = 5t + 0.4t^2. Here d = 165, and we solve that t = 15, and v = 14.
2. Let a denote the deceleration of the train. Let v denote the initial velocity of the train. Then we have v*20 - 0.5*a*20^2 = 0.25 (1), and v - 20*a = v - 10 (2). Solving equations (1) and (2), we get a = 0.5, v = 0.5125.
3. Let a denote the cyclist's acceleration. Let v denote his initial velocity. We have the following: 100*v + 0.5*a*100^2 = 1000 (1), 80*(v + 100*a) + 0.5*a*80^2 = 1000 (2). Solving equations (1) and (2), we get a = 0.03, v = 8.61.
2. Let a denote the deceleration of the train. Let v denote the initial velocity of the train. Then we have v*20 - 0.5*a*20^2 = 0.25 (1), and v - 20*a = v - 10 (2). Solving equations (1) and (2), we get a = 0.5, v = 0.5125.
3. Let a denote the cyclist's acceleration. Let v denote his initial velocity. We have the following: 100*v + 0.5*a*100^2 = 1000 (1), 80*(v + 100*a) + 0.5*a*80^2 = 1000 (2). Solving equations (1) and (2), we get a = 0.03, v = 8.61.
1.A cyclist comes to the top of a hill 165 metres long trave
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