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| Math Cadet |
Joined: Wed, 3 Dec 2008 11:17:17 UTC Posts: 6
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Car A is travelling on a highway at a constant speed of 100km/h and is 120m from the entrance of an access ramp.
100 km/h = 27.8 m/s
Car B enters the acceleration lane at that point at a speed of 25 km/hr
25 km/h = 6.95 m/s.
Car B accelerates uniformly and enters the main traffic lane after travelling 70 m in 5s.
It then continues to accelerate at the same rate until it reaches a speed of 100km/hr. which it then maintains.
Determine the final distance between the two cars.
CAR B
Ave, speed on the ramp
70 m / 5 s = 14 m/s
Final speed on the ramp
( 6.95 m/s + final speed) / 2 = 14 m/s
final speed = 21.05 m/s
acceleration on the ramp
(21.05 m/s – 6.95 m/s) / 5s = 2.82 m/s/s
time taken to reach top speed
(27.8 m/s – 6.95 m/s) / 2.82 m/s/s = 7.4 s
average speed
(6.95 m/s + 27.8 m/s) / 2 = 17.38 m/s
distance travelled
17.38 m/s x 7.4 s = 128.6 m
CAR A
Distance travelled
27.8 m/s x 7.4 s = 205.7 m
distance of A from B
120 m + 128.6 m – 205.7 m = 42.9 m
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