Two cars are at the start lines of two roads at right angles to each other and p
ID: 1820039 • Letter: T
Question
Two cars are at the start lines of two roads at right angles to each other and pointing towards the cross-road intersection 500 m away. Car A starts from rest with uniform acceleration of 0-5 m/s2 and accelerates for 20 seconds. It then continues at the speed reached. Car B starts 5s later and 20m further back from the start distance of car A with an acceleration of 0.7 m/s 3. calculate, with respect to start time of ear A:- The time it takes for B and A to be at the same distance from the cross-roads. The distance from the cross-roads at the above time. The average velocity of car A at this distance. Car B stops accelerating at this point continues at the speed reached What is the shortest distance between the two cars as they pass the cross roads.Explanation / Answer
a) from Car A: acceleration = 0.5 m/s^2 time = 20 seconds initial velocity = 0 m/s final velocity: v= u + at = 0 + 0.5(20) = 10 m/s distance travelled during acceleration: d = t(at + 2u)/2 = 20(0.5(20) +2*0)/2 = 100 m distance travel when reach constant velocity: d = 10t m for Car B, 100 + 10(t-20) + 20 = (t-5)(0.7(t-5 + 2*0)/2 100 + 10t - 200 + 20 = (t^2 - 10t + 25)(0.35) 10t - 80 = 0.35t^2 - 3.5t + 8.75 0.35t^2 -13.5t + 88.75 = 0 (-17.8461 + 0.591608t) (-4.97308 + 0.591608t) = 0 t = 8.406 or 30.165 seconds b) i) when t = 8.406 seconds, d = t(at + 2u)/2 = 8.406(0.5(8.406) +2*0)/2 = 17.665 m distance from cross road: 500 - 17.665 = 482.335 m when t = 30.165 seconds, 30.165 - 20 =10.165 seconds distance = 10.165 * 10 + 100 = 101.65 + 100 = 201.65 m distance from cross road: 500 - 201.65 = 298.35 m ii) when t = 8.406 seconds, average velocity = 17.665/8.406 = 2.101 m/s when t = 30.165 seconds, average velocity = 201.65/30.165 = 6.685 m/s c) Car B stop accelerating at t = 8.406 seconds: v= u + at = 0 + 0.7(8.406-5) = 2.384 m/s time taken for car A to reach cross road: (500-100)/10 = 40 seconds ***this is after Car A reach constant velocity. time taken for car B to be in constant velocity when car A reach cross road: 40 + 15 - (8.406 - 5) = 51.594 seconds distance travel by car B: distance travel during acceleration + distance travel during constant velocity = (8.406-5)(0.7(8.406-5) +2*0)/2 + [0 + 0.7(8.406-5)]* 51.594 = (3.406)(0.7*3.406)/2 + (0.7*3.406)(51.594) = 127.071 m shortest distance = 500 + 20 - 127.071 = 392.93 m ---------------------------------------------------------------------------------------- Car B stop accelerating at t = 30.165 seconds: v= u + at = 0 + 0.7(30.165-5) = 17.616 m/s distance travelled during acceleration: d = (30.165-5)(0.7(30.165-5) +2*0)/2 = 221.647 m time taken for car B to reach cross road: (500 + 20 - 221.647)/17.616 = 16.936 seconds time taken for car A to be in constant velocity when car B reach cross road: (30.165 - 20) + 16.936 = 27.101 seconds distance travel by car A: distance travel during acceleration + distance travel during constant velocity = 100 + 10* 27.101 = 100 + 271.01 = 371.01 m shortest distance = 500 - 311.01 = 188.99 m
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