Two cars 1 and 2 starting from rest are moving with speeds V1 and V2 m/s (V1>V2).Car 2 is ahead of car 1 by'S' metres when the driver of car'1' sees car'2'.What minimum retardation should be given to car'1' avoid collision.
Answers
Answered by
7
1 answer · Physics
Best Answer
1.) Since we are given a position function f(t) = 12t + 3t^2 - 2t^3, we can use the derivative of this position vector to find its velocity at point t = 0 (or, the start). f'(t) = 12+ 6t - 6t^2. Plugging in t = 0, we get that f'(0) = 12 + 0 - 0 = 12, so the car will have a velocity of 12m/s at point t = 0. I realize you've requested no use of calculus, but there is no other way to determine this problem: velocity is the derivative of position, and dividing by t to get s/t (velocity) would be equivalent to dividing by 0.
2.) Using the kinematic equation vf^2 = vi^2 + 2ax, we want to know vf, given that vi = 0, a = 9.8m/s^2, and x = 50m. Thus:
vf^2 = 2(9.8)(50m) = 980m^2/s^2
vf = Sqrt(980m^2/s^2) = 31.3m/s.
Next, we know that he decelerates (I despise physics teachers who use decelerate instead of negative acceleration) at a rate of 2m/s^2 until he reaches a speed of 3m/s. Subtracting 3 from 31.3, we know that the acceleration applies until 28.3m/s becomes 0m/s. Using the same formula as before, we can use vf = 0, vi = 28.3, and a = -2 to determine x, like so:
0 = (28.3)^2 - 2(2)x -> x = ((28.3)^2)/4 = 200.223m.
Adding the 50 original feet, we get that he bailed out at approx. 250m.
Best Answer
1.) Since we are given a position function f(t) = 12t + 3t^2 - 2t^3, we can use the derivative of this position vector to find its velocity at point t = 0 (or, the start). f'(t) = 12+ 6t - 6t^2. Plugging in t = 0, we get that f'(0) = 12 + 0 - 0 = 12, so the car will have a velocity of 12m/s at point t = 0. I realize you've requested no use of calculus, but there is no other way to determine this problem: velocity is the derivative of position, and dividing by t to get s/t (velocity) would be equivalent to dividing by 0.
2.) Using the kinematic equation vf^2 = vi^2 + 2ax, we want to know vf, given that vi = 0, a = 9.8m/s^2, and x = 50m. Thus:
vf^2 = 2(9.8)(50m) = 980m^2/s^2
vf = Sqrt(980m^2/s^2) = 31.3m/s.
Next, we know that he decelerates (I despise physics teachers who use decelerate instead of negative acceleration) at a rate of 2m/s^2 until he reaches a speed of 3m/s. Subtracting 3 from 31.3, we know that the acceleration applies until 28.3m/s becomes 0m/s. Using the same formula as before, we can use vf = 0, vi = 28.3, and a = -2 to determine x, like so:
0 = (28.3)^2 - 2(2)x -> x = ((28.3)^2)/4 = 200.223m.
Adding the 50 original feet, we get that he bailed out at approx. 250m.
Answered by
15
Answer:
(V1-V2)^2÷2S
Explanation:
anyone can solve this question using formulas of kinematics
Similar questions