Two cars start from rest at a red stop light. When the light turns green, both c

Question

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.4 m/s2 for 4.1 seconds. It then continues at a constant speed for 6.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 140.47 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

What is the acceleration of the blue car once the brakes are applied?

What is the total time the blue car is moving?

What is the acceleration of the yellow car?

Explanation / Answer

:  Blue car:

Velocity after acceleration = (at), = 3.4 x 4.1, =13.94m/sec.

Distance covered during acceleration = 1/2 (vt), = 1/2 (13.94 x 4.1), = 28.57m.
It then covers another (13.94 x 6.1) =85.03m. before braking.
(85.03+ 28.57) = 113.6m.
Did you add the "approx.302.2m" in there? I'd wonder why all the other info was given, if they (the question authors) then give a different figure? I'll use 304.92.....
It covers (140.47 - 85.03) = 55.44 metres while braking. It's doing 13.94m/sec., accelerating to 0.
1) Acceleration = (v^2/2d), = (113.6/57.14) = 1.98m/sec^2.
2) Time to stop = sqrt. (2d/a), = sqrt. (57.14/1.98) = 5.37s.
Total time = (5.37+ 6.1 + 4.1) = 15.57s.

Total distance blue car moves =140.47m. The yellow car has been accelerating the entire time, 15.57s.
3) Yellow car acceleration = (2d/t^2), = (227.2/ 242.4), = .93m/sec^2.

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