Particle's Velocity in t: -kt^3 + c

[Cbray]

Homework Statement

A particle moves in a straight line with acceleration which is inversely proportional to t³ , where t is the time. The particle has a velocity of 3ms^-1 when t=1 and its velocity approaches a limiting value of 5ms^-1 . Find an expression for its velocity at time t.

Homework Equations

a=-kt^3

The Attempt at a Solution

*****ROOKIE MISTAKE SHOULD BE -2/t^2 + d (last equation)***

 

[agent_509]

I won't answer the question directly for you, but this is just a basic algebra problem, remember that when y is proportional to x, you use the formula y=kx where k is some constant, inversly proportional to x and you have y = k/x. Once you know how y is proportional to x, and you have test values for y and x, you can plug them in and solve for k.

 

[Cbray]

I won't answer the question directly for you, but this is just a basic algebra problem, remember that when y is proportional to x, you use the formula y=kx where k is some constant, inversly proportional to x and you have y = k/x. Once you know how y is proportional to x, and you have test values for y and x, you can plug them in and solve for k.

I did some more and got stuck again, do you mind looking at it?

 

[agent_509]

I'm not entirely sure what you did there here's what would be a good idea:

∫ dv/dt dt = k∫1/t^2 dt

v= -k/2t^2 + d

now you know that when t=1 v=3, and you also know that the limiting velocity (when t→∞) is 5, so when t→∞, d=v, and v= 5, so d = 5, so now you have the formula

v=-k/2t^2 + 5, and you know that when t=1, v=3.

so just plug it in and solve for k