integration(3)

Larry

2011-02-21 3:30 am

Initially, a particle moves with a velocity of 24 m/s along a straight line and its acceleration after t seconds is (8-15t^2) m/s^2. It stops instantaneously and then reverses its direction of motion.a) show that the particle stops instantaneously at t=2bi) find the distance travelled by the particle in first 2 secondsbii) dinf the distance travelled by the particle in first 4 seconds

更新1:

bii) ans is 248m

回答 (1)

用戶25975

2011-02-21 4:07 am

✔ 最佳答案

a) The velocity is:

∫ (8 - 15t2) dt

= 8t - 5t3 + C

With v = 24 when t = 0, we have C = 24

Hence v = 8t - 5t3 + 24

When t = 2, v = 16 - 40 + 24 = 0

Hence the particle stops inst. at t = 2

b i) ∫ (t = 0 → 2) v dt

= ∫ (t = 0 → 2) (8t - 5t3 + 24) dt

= [4t2 - 5t4/4 + 24t] (t = 0 → 2)

= 44 m

b ii) ∫ (t = 2 → 4) v dt

= [4t2 - 5t4/4 + 24t] (t = 2 → 4)

= - 160 m

Hence the distance travelled by the particle in first 4 secs = 44 + 160 = 204 m

2011-02-20 20:17:39 補充：
Some typo mistake for b ii)

[4t^2 - 5t^4/4 + 24t] (t = 2 → 4) = -204 m

Hence total distance = 204 + 44 = 248 m

參考: 原創答案

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