數學問題 40分

Question 1.
The displacement is given by s = 8t - t^2
Differentiate with respect to t
ds/dt = 8 -2t (This is instantaneous velocity)
When t = 0s
ds/dt = 8 -2t
d^2s.dt^2 = 8 m/s^2 (This is instantaneous acceleration)
When t = 0 s
ds/dt = 8 -2(0)
ds/dt = 8 m/s
When t = 0, d^2s.dt^2 = 8 m/s^2

(a)Its initial velocity = 8 m/s and initial acceleration = 8 m/s^2

When the object change its direction of motion, the velocity change sign
ds/dt = 8 -2t
0 = 8 – 2t
2t = 8
t = 8/2 = 4 s
From 0 to 4 s, ds/dt is positive; from 4s onward, ds/dt is negative. 

(b) The object will change its direction of motion when t = 4s


Question 2.
r^2 = x^2 + y^2 where r is the length of the rope, x is the distance of the boat from the shore and y is the height of pulley above the boat
Do an implicit differentiation, differentiate with respect to time t
2r dr/dt = 2xdx/dt + 2y dy/dt
r dr/dt = xdx/dt + y dy/dt
When the boat is 5 m from the shore,
x = 5m, y = 2m, r = Sq. root of(x^2 + y^2) = Sq. root (5^2 + 2^2) = 5.3852 m
dy/dt = 0 because the pulley is not moving (always 2m above the boat)
r dr/dt = xdx/dt + 0
dr/dt = 0.5 m/s, dx/dt = ?, r = 5.3852 m, x = 5 m, We have to find dx/dt
(5.3852 m)(0.5 m/s) = (5m) dx/dt
dx/dt = (5.3852 m)(0.5 m/s)/(5m) = 0.538 m/s

The boat moving at the rate of 0.538 m/s when the boat is 5 m from the shore.