math question

2007-10-19 6:39 pm
a particle moves along the x-axis ins such a way that its postion s at time t is
s(t)= 3t^4-16t^3+24t^2, for -5 is less than and equal to t, t is less than and equal to 5
(a) determine the velocity and acceleration of the particle at time t.
(b) at what values of t is the particle at rest?
(c) at what values of t does the particle change direstion?
(d) what is the velocity when the acceleration is zero for the first time.
(e) determine the distance travelled from t=0 to t=5.

回答 (2)

2007-10-19 6:59 pm
參考: Myself~~~
2007-10-19 7:39 pm
Solution to part (d) given above is not correct.

(d)
3t^2-8t+4=0
(3t-2)(t- 2)=0
t=2/3 or t=2
For t = 2/3,
v(2/3)
=12(2/3)^3-48(1/3)^2+48(1/3)
=32/9 - 16/3 +16
=128/9
參考: My Maths knowledge


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