超難的數學Word Problem!!!!!!!! 請高手入

(1) Reasoning:

By definition, the velocity of an object is the rate change of distance travelled, denoted s, mathematically, we have

v = ds/dt.

We are given : s(t) = 4.9t^2.

(2) Consider our case:

v = ds/dt = s'(t)

(3) Differentiate: by the constant multiplication and power rule, we have

s'(t) = 4.9(2)t = 9.8t

(4) Substitute: t = 3, we have = 9.8(3) = 29.4


2012-01-27 05:26:41 補充：
velocity should be the rate change of displacement. but for a easier understanding in a free falling object, distance is equivalent to displacement in our case.

2012-01-27 05:30:02 補充：
We can also deduce the velocity by linear motion equations in physics, where v^2 = u^2 + 2as.

a = 9.8 m/s^2 . It is the approximation of the acceleration due to gravity, usually say g.

Here u = 0 (initial velocity) for an object to be released from rest.
We get v (final velocity) = 29.4145... m/s

s = 4.9t^2v = ds/dt = 9.8tSo, v(3) = 9.8 * 3 = 29.4The velocity of the object after 3 seconds is 29.4 m/s

2012-01-24 18:07:36 補充：
上面排版不清楚

s = 4.9t^2

v = ds/dt = 9.8t

So, v(3) = 9.8 * 3 = 29.4

The velocity of the object after 3 seconds is 29.4 m/s