Variable acceleration, time, distance

[Markisen]

Homework Statement

A particle moves along a straight line. At t=0 the acceleration is k(t+2), initial velocity = 0 and x=5m.
a) Write down the position of the particle as a function of time.
b) At what point in time does the particle reach the velocity 16m/s, and where is the particle at this time?

Homework Equations

[tex]a=k(t+2)=kt+2k[/tex]
[tex]v=\int a\textit{dt}=(kt^2)/2+2kt+C_1[/tex]
[tex]x=\int v\textit{dt}=(kt^3)/6+kt^2+C_2[/tex]

The Attempt at a Solution

I think I got part a) With the above equations that I calculated and with the initial values inserted I get: [tex]x=(kt^3)/6+kt^2+5[/tex]

for part b I'm more unsure that I'm correct. Entering the velocity as 16m/s I get the equation: [tex]16=(kt^2)/2+2kt[/tex] which when calculated gets the time. [tex]t=-2+4\sqrt{1+{64/k}^2}[/tex] and then this time is inserted into [tex]x=(kt^3)/6+kt^2+5[/tex] to calculate the distance. This gives me [tex]1/6\,k \left( -2+4
\,\sqrt {1+64\,{k}^{-2}} \right) ^{3}+k \left( -2+4\,\sqrt {1+64\,{k}^{-2}} \right) ^{2}+5 [/tex] Seems a bit to complicated to be right? Or is it right and I'm just confused

Thanks for the help!

 

[CompuChip]

Note that
[tex]x=\int v\textit{dt}=(kt^3)/6+kt^2+C_1 t + C_2,[/tex]
you forgot to integrate the C1 term.

Finally, you can probably simplify your answer a bit but your way of calculating it looks correct. I am afraid that you are just confused and it is just an ugly answer

 

[nlightner]

Your solution for part a) is correct. However, for part b), your equations are not correct. The time at which the particle reaches a velocity of 16m/s should be when v=16, not when a=16. So, the correct equation to use would be v=kt+2k, and solving for t will give you the time at which the particle reaches a velocity of 16m/s. Then, to find the position of the particle at this time, you can plug in this value of t into your equation for x to get the distance. Your final answer should not involve k, as it is a constant and will cancel out in the calculations. The final answer should be a numerical value for the distance.