Physics Motion (10pts)

2011-02-21 2:53 am
1) Jason skis down a mountain from rest (FigH) The first section of the path has a smaller slope and Jason accelerates at 0.1ms- 2 After skiing 500m Jason reaches the second section of path, which has a greater slope and is 800m long. His acceleration increases to 0.5ms-2


圖片參考:http://imgcld.yimg.com/8/n/HA07083779/o/701102200127313873382630.jpg
Fig H

a) What is Jason's speed when he has finished the first 500m?

b)find the total time taken for Jason to travel downhill.




2) Fireworks are best displayed when they explode at their top projection points.Two fireworks X and Y are designed to explode at different heights simultaneously.

圖片參考:http://imgcld.yimg.com/8/n/HA07083779/o/701102200127313873382631.jpg
Fig B

a) What is the velocity of firework X when it is fired?

b) How long does it take for firework X to reach that height?

c)using the results in a) and b) ,suggest an arrangement such that firework Y explodes simultaneously with firework X.

回答 (1)

2011-02-21 3:27 am
✔ 最佳答案
1.(a) Consider the first section of slope,
u = 0 m/s, a = 0.1 m/s2, s = 500 m, v =?Use equation: v^2 = u^2 + 2asv^2 = 2 x 0.1 x 500 (m/s)^2
i.e. v = 10 m/s

Consider the second section of slope,
u = 10 m/s, a = 0.5 m/s2, s = 800 m, v =?hence, v^2 = 10^2 + 2 x 0.5 x 800 (m/s)^2
v = 30 m/s

(b) Apply equation v = u +atTime taken for the 1st section = 10/0.1 s = 100 sTime taken for the 2bd section = (30-10)/0.5 s = 40 sHence, total time taken = (100 + 40) s = 140 s
2. (a) Use equation of motion: v^2 = u^2 + 2as
with v = 0 m/s, a = -10 m/s2, s = 200 m, u = ?hence, 0 = u^2 + 2.(-10).(200)solve for u gives u = 63.25 m/s(b) Use equation: v = u + at0 = 63.25 + (-10)tt = 6.325 s
(c) Use equation: s = ut + (1/2)at^2
with s = 130 m, t = 6.325 s, a = -10 m/s2, u = ?
hence, 130 = 6.325u + (1/2).(-10).(6.325)^2
u = 52.2 m/s
Hence, to have the two fireworks eplode simultaneously, firework Y needs to have an initial speed of 52.2 m/s




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