One dimension motion (1 Viewer)

[YBK]

Hey, could anyone please help me out with this question:

A particle moves in a straight line with acceleration wihich is inversely proportional to t^3, where t is the time. The particle has a velocity of 3m/s when t=1 and its velocity approaches a limiting value of 5m/s. Find an expression for its veloctiy at time t.


Thanks!

 

[Riviet]

I think this is how you do it:

a=dv/dt=k/t³, k is a constant
Integrate both sides with respect to t,
v=-kt^-2/2+C
When t=1 and v=3,
3=-k/2+C
C=3+k/2
.'. v=-kt^-2/2+3+k/2
lim (-k/2t²+3+k/2) = 3+k/2 = 5, since velocity approaches 5m/s.
t->infinity
.'. k=4
.'. v=-2/t² + 5

 

[pLuvia]

Told acceleration wihich is inversely proportional to t^3
a=k/t³, where k is a constant
dv/dt=k/t³
v=-k/2t²+C
When t=1 v=3, C=3+k/2
v=-k/2t²+3+k/2

velocity approaches a limiting value of 5m/s
This means when v_max=5
This occurs when a=0
0=-k/t³
t-->infinity
When t-->infinity v=5
5=-k/2(t-->infinity)+3+k/2
2=k/2
k=4

v=-2/t²+5

 

[YBK]

cool, thanks a lot!