急15分 Mechanics -Linear Motion1－bxzbthf的部落格

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急15分 Mechanics -Linear Motion1

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**4. Before taking off, anaircraft travelling with a constant accerleraion makes a run on a field of 1800m in 12 second from rest. Determine (a) the acceleration, (b) the velocity withwhich it takes off, (c) the distance covered in the 1st and the 12thseconds.請提供詳解、算式及答案 **8. A tower is... 顯示更多 **4. Before taking off, anaircraft travelling with a constant accerleraion makes a run on a field of 1800m in 12 second from rest. Determine (a) the acceleration, (b) the velocity withwhich it takes off, (c) the distance covered in the 1st and the 12thseconds.請提供詳解、算式及答案 **8. A tower is 100m high. One body is dropped from the top of the tower, andat the same instant another body is projected vertically upwards from thebottom, and they meet half-way up. Find (a) the initial velocity of theprojected body, (b) its velocity when it meets the descending body. 請提供詳解及算式[Answer: (a) 31.32 m s^-1 / (b)0.006 m s ^-1] **9. A rocket leaves its padvertically with an acceleration of 7.5 m s^-2 which remains constant until thefuel is exhaused after 8 s. It then continues to travel freely in the verticaldirection. Find (a) the total time taken for the rocket to reach its highestpoint and (b) the altitude acheived. 請提供詳解及算式[Answer: (a) 14.1 s / (b) 424m] **10. A car accelerates fromrest, reaching a velocity of 8 m s^-1after 20 s. It is then decelerated at 0.4m s^-2 until its velocity drops to 6 m s^-1. At this moment, the braking forceis increased and the car comes to a stop under a deceleration of 1.2 ms^-2.Calculate (a) the total distance covered while the car is in motion and (b) itsaverage velocity. 請提供詳解及算式[Answer: (a) 130 m / (b) 4 1/3m s^-1]

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4a) Applying s = ut + at2/2 with u = 0, s = 1800 and t = 12: 1800 = 72a a = 25 m/s2 b) Applying v = u + at with u = 0, a = 25 and t = 12: v = 25 x 12 = 300 m/s c) Applying s = ut + at2/2 with u = 0, a = 25 and t = 1: s = 25/2 = 12.5 m (Distance covered in 1st sec) With u = 0, a = 25 and t = 11, s = 1512.5 m (Distance covered in first 11 secs) Hence distance covered in the 12th sec = 1800 - 1512.5 = 287.5 m 8a) Time needed for the dropped body to travel half-way 50 m is: s = ut + at2/2 with u = 0, a = 9.8 and s = 50 50 = 4.9t2 t = 3.19 s For the projected body, applying s = ut + at2/2 with t = 3.19, a = -10 and s = 50 50 = 3.19u - 5 x 3.192 u = 31.32 m/s b) With v = u + at: v = 31.32 - 9.8 x 3.19 = 0.06 m/s 9a) After the fuel is exhausted, the rocket's velocity is: v = 7.5 x 8 = 60 m/s upward After that the rocket is under free fall with a = 9.8 m/s2 So further time needed for it to reach highest point = 60/9.8 = 6.1 s Total time = 8 + 6.1 = 14.1 s b) Before the fuel is exhausted: s = ut + at2/2 with t = 8, a = 7.5 and u = 0: s = 240 m After the fuel is exhausted: s = ut + at2/2 with t = 6.1, a = -9.8 and u = 60: s = 184 m Total altitude = 424 m 10a) Initial acc. = 8/20 = 0.4 m/s2 s = ut + at2/2 with t = 20, a = 0.4 and u = 0: s = 80 m In the first braking stage: v2 - u2 = 2as with v = 6, a = -0.4 and u = 8: -28 = -0.8s s = 35 m In the second braking stage: v2 - u2 = 2as with v = 0, a = -1.2 and u = 6: -36 = -2.4s s = 15 m Total distance = 80 + 35 + 15 = 130 m b) Time for first braking stage = (8 - 6)/0.4 = 5 s Time for second braking stage = (6 - 0)/1.2 = 5 s Total time used = 40 s Average velocity = 130/40 = 13/4 m/s

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