PHY mechanics

無最佳解答

(a) Use equation of motion: s = ut + (1/2)a.t^2, and take the origin at the position of car B at time t = 0 s
For car A; S(a) = 4t + 7
For car B: S(b) = 2t + (1/2)(2)t^2

hence, S(a) - S(b) = 4t + 7 - 2t - t^2
Let x be the distance between cars A and B,
x = 2t - t^2 +7
dx/dt = 2 - 2t
For max value of x, dx/dt = 0
hence, 2 - 2t = 0
i.e. t = 1 s

Max distance between the two cars = (2 x 1 - 1^2 + 7) m = 8 m

(b) Let T be the time required
When car B catches car A, x = 0
hence, 2T - T^2 + 7 = 0
or T^2 - 2T - 7 = 0
solve for T gives T = 3.828 s

2010-03-10 23:01:39 補充：
In part (a), you could use the following method without using calculus.
The max distance occurs at the time when the speed of car B equals to that of car A. After that, car B is faster than car A, and their separation shortens....

2010-03-10 23:08:12 補充：
(contd)...speed of car A = 4 m/s, speed of car B =2+2t (use v = u+at)
when they have the same speed, 4 = 2+2t, hence, 2t=2, or t = 1s
Then use x = 2t - t^2 +7 to calculate max distance between 2 cars