A car starts from rest at point A and accelerates at the rate of 0.8 m/s2 until it reaches a speed of 12 m/s. It then proceeds at 12 m/s until the brakes are applied. It comes to rest at point B, 42 m beyond the point where the brakes are applied. Assuming uniform deceleration and knowing the distance between A and B is 300 m, determine the time required for the car to travel from A to B
I'm unsure how to finish it. I got a displacement of 90 meters and 15 seconds for the acceleration period, but I don't know how to calculate the displacement and time for the constant speed or deceleration period.
Help, much apperciated
I'm unsure how to finish it. I got a displacement of 90 meters and 15 seconds for the acceleration period, but I don't know how to calculate the displacement and time for the constant speed or deceleration period.
Help, much apperciated
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Assume constant deceleration over the last 42m (given). Find a, given 12m/s and 42m. Then you should be able to find time. Displ. for middle period can be found by subtracting the other displacements from the total distance.
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time for acceleration = 12 / 0.8 = 15s
Distance covered during acceleration = 1/2 . a . t^2 = 0.4 . 15^2 = 90 m
Decelerating distance = 42 m
Deceleration = v^2 / 2 . d = 12^2 / 2 . 42 = 1.71 m/s/s
Time for deceleration = v / a = 12 / 1.71 = 7 s
Steady speed distance = 300 – 90 – 42 = 168 m
Time to do 168 m at 12 m/s = 14 s
Total time taken = 14 + 7 + 15 = 36 s
Distance covered during acceleration = 1/2 . a . t^2 = 0.4 . 15^2 = 90 m
Decelerating distance = 42 m
Deceleration = v^2 / 2 . d = 12^2 / 2 . 42 = 1.71 m/s/s
Time for deceleration = v / a = 12 / 1.71 = 7 s
Steady speed distance = 300 – 90 – 42 = 168 m
Time to do 168 m at 12 m/s = 14 s
Total time taken = 14 + 7 + 15 = 36 s