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Road Dragster
Acceleration homework problems?
I just finished most of my fitness, but there are some issues that do not understand. I tried the figure again and again, but it makes no sense. Can someone please explain step by step for each problem. best answer will get points. 1) A starting dragster accelerates from rest to 49 m / s ² How fast he goes when he has run 325 m? 2) A car moving at 12 m / s, and coasts up a hill with an acceleration uniform of -1.6 m / s ² a) If the car is 55 m / s, how many meters will travel before it stops? B) How many meters is needed to stop a car is twice as fast? 3) A car is traveling 20.0 m / s when the driver sees a child standing on the road. She takes 0.80 s to react then steps on the brakes and slows to 7.0 m / s ² Since fat is the car go before it stops?
[[1]] Find the variables. V (i) = 0 m / s (which from rest), a = 49 m / (s ^ 2) d = 325 m. Use this equation of motion to find the first time: d = V (i) t + 0.5 at ^ 2, V (i) = 0 for d = 0.5 a ^ 2. Alternate, 325 = 0.5 (49m / (s ^ 2)) (t ^ 2), so that 650 = 49t ^ 2, divided by t ^ 2 = 49 13.27 s – t> = 3.64s . Now find the terminal velocity, using the equation of motion V (f) = V (i) + a, V (i) = 0 for the replacing, V (f) = (49 m / (s ^ 2)) (3.64 s) = 178.36 m / s. [[2]] a = -1.6 m / (s ^ 2) (Please tell me if I read this wrong, I do not see the initial speed of 12 m / s are used in parts A and B in any way ) V. [A] (i) = 55 m / s, V (f) = 0 m / s (as it is reduced to a stop). Find distance using the equation of motion V (f) ^ 2 = V (i) ^ 2 + 2ad, so that 0 = 3025 + 2 (-1.6) d -> -3025 =- 3.2d, split, d = 945.3 m. [B] V (i) = 2 * 55 m / s = 110 m / s, V (f) = 0. Use the same as used last time, giving -12100 =- 3.2d, dividing, d = 3781.25 m [[3]] V (i) = 20 m / s, a = -7 m / (s ^ 2). First find the distance before she reacts, using no acceleration distance = rate x time: d = 20 m / s * 0.8 s = 16 m. Now find the time it takes to reach a stop, with V (f) = 0: V (f) = V (i) + a, 0 = 20 +-7t -> -20 =-7t, t = 2.86s. Now find the distance. D = V (i) t + 0.5 at ^ 2 d = 20 (2.86) + 0.5 (-7) (2.86 ^ 2) = 57.2 m – m m = 28.6286 28.5714 Add this to the reaction time distance, d = 28.5714 m + 16 m = 44.5714 m.
