力學...mechanics

1) A car undergoes unform deceleration along a straight road.Its velocity decreases from 40ms^-1 to 30ms^-1 after travelling a distance of 100m. How much more time will it take to come to rest?
2aS = v^2 -u^2
v = 30ms^-1; u = 40ms^-1 and S = 100m
Deceleration a = (30^2 - 40^2)/ (2 * 10) = (900 - 1600)/20 = -35ms^-2
at = v - u
a = -35ms^-2; v = 0 ms^-1
Time to come rest t = (0 - 30) / -35 = 0.857s

2)An object is throw vertically upwards with a speed of 15ms^-1 from a cliff.It takes 10s for the object to reach the ground.What is the height of the cliff above the ground??
S = ut + 1/2 * at^2
u = 15ms^-1; t = 10s; a = -9.81ms^-2 (downward)
Height of the cliff S = 15 * 10 + 1/2 * -9.81 * 10^2 = 150 - 490.5 = -340.5m

2006-12-09 01:04:09 補充：
1) 2aS = v^2 -u^2v = 30ms^-1; u = 40ms^-1 and S = 100mDeceleration a = (30^2 - 40^2)/ (2 * 100) = (900 - 1600)/20 = -3.5ms^-2at = v - ua = -3.5ms^-2; v = 0 ms^-1Time to come rest t = (0 - 30) / -3.5 = 8.57s

(1) v^2 - u^2 = 2as
30^2 - 40^2 = 2*100*a
a = -3.5 ms^-2
Use v = u + at to find the time,
v = 0, u = 30, a = -3.5
3.5t = 30
t = 8.57 s
So the car will take 8.57seconds more to come to rest.

(2) Let h be the height of the cliff
Take the downward direction of the vector quantities to be positive,
use s = ut + 1/2*at^2
h = (-15)(10) + 0.5*(10)*(10)^2
= 350 m
So the height of the cliff above the ground is 350 m.