- #1

Markisen

- 1

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## 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!