Question

the velocity of a particle is given by v=16t^2i+4t^3j+(5t+2k) m/s, where t is in seconds. if the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. also what is the x,y,z coordinate position of the particle at this instant

Answer 1

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.

$v_{x}=16t^{2} \\v_{y}=4t^{3}\\v_{z}=5t+2$

acceleration is a derivative of velocity with respect to time.

$a_{x}=\frac{d}{dt} v_{x}=\frac{d}{dt}[16t^{2}]=32t\\a_{y}=\frac{d}{dt} v_{y}=\frac{d}{dt}[4t^{3}]=12t^{2} \\a_{z}=\frac{d}{dt} v_{z}=\frac{d}{dt}[5t+2]=5$

evaluate acceleration at 2 seconds

$a_{x} =32*2=64m/s^{2}\\ a_{y} =12*2^{2} =48m/s^{2}\\a_{z} =5m/s^{2}$

the magnitude of the acceleration is the square root of the sum of the square of each component of the acceleration.

$=\sqrt{a_{x}^2 +a_{y}^2+a_{z} ^2 } \\=\sqrt{64^2 +48^2+5 ^2 }\\=80.16m/s^2$

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.

$x=\int\limits {v_{x} } \, dx=\int\limits^2_0 {{16t^2} \, dt=42.7m\\\\y=\int\limits {v_{y} } \, dx=\int\limits^2_0 {{4t^3} \, dt=16m\\\\\\\\\\z=\int\limits {v_{z} } \, dx=\int\limits^2_0 {{5t+2} \, dt=14m$