Application of diff and int

(a)

dS/dt = 7t+5

S = ∫ (7t+5) dt [ t from 0 to 4]

S = 7t² + 5t | [ t from 0 to 4]

S = 7(4)² + 5(4) = 132 m

(b)

Distance travel by car A from 0 to T
= 7T² + 5T | [ t from 0 to T] .... (1)

Distance travel by car B from 5 to T
S = ∫ (9t-42) dt [ t from 5 to T]

S = 9t² - 42t | [ t from 5 to T]

S = 9T² - 42T -[9(5)² - 42(5)]

S = 9T² - 42T -15

When distance travel by car A = When distance travel by car B

7T² + 5T = 9T² - 42T -15

2T² - 47T -15 = 0

T = 23.8 s or -0.315 s(reject)

(c)

S = 7t² + 5t | [ t from 0 to 23.8]

= 7(23.8)² + 5(23.8)

= 4084 m

2013-12-29 23:50:37 補充：
注意 :

Vb = 9t - 42

when t = 23.8

Vb = 172.2 m/s

= 620 km/h

時速 620 公里的車實在太快啦！所以，這條只是數學題，並不真實。

唔怪得 二十幾秒會行咗四公里咁多啦！