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A particle moving with v = (30m/s)i undergoes an acceleration a = [5 m/s^2 +(3m/s^s)t^3]i + [45^2 - (1m/s^2)t^2]j. What are the particle position...

Hello!


There are some misprints in formulas here, I suppose that the acceleration is


`a=a(t)=(5+3t^3)i+(45-t^2)j,`


where t is for time in seconds.



By definition, acceleration is the derivative of speed by time, `a(t)=v'(t).` Therefore `v(t)=v_0+int_0^t a(x) dx.` The same way the position is `s(t)=s_0+int_0^t v(x) dx.` It is given that `v_0=30 i` and `s_0=0.`



Integrating once, obtain


`v(t)=(30+5t+3/4 t^4) i + (45t-1/3 t^3) j,`


integrating twice obtain


`s(t)=(30t+5/2 t^2 + 3/20 t^5) i + (45/2 t^2 - 1/12 t^4) j.`



Now we have to substitute `t=4` into these equalities. So


`v(4)=(30+20+192)i+(180-64/3)j=242i+(158+2/3)j` (in m/s)


and


`s(4)=(120+40+153.6)i+(360-21-1/3)j=313.6i+(338+1/3)j` (in meters).


These are the answers. If some functions are misprinted, you can integrate the correct functions.

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