Answer:
80.16 m/s^2
at t=2 s
x=42.3 m
y=16 m
z=14 m
Explanation:
solution
The x,y,z components of the velocity are donated by the i,j,k vectors.
[tex]v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2[/tex]
acceleration is a derivative of velocity with respect to time.
[tex]a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5[/tex]
evaluate acceleration at 2 seconds
[tex]a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}[/tex]
the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.
[tex]=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2[/tex]
position is the integral of velocity with respect to time position at a time can be found by taking by taking the definite intergral of each component.
[tex]x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m\\\\[/tex]
