Acceleration, Position and Velocity Vectors?

Question

Please help me with the following question.

A particle moves such that its velocity is:
V(t) = (12t^3 + 2t)i - (4t)j + (6t^2 - 4)k
Find its acceleration and position, given that the particle starts at i-j. Is there a time at which the particle is at rest?

I have differentiated the above equation to get an expression for acceleration and I have integrated the above equation to get an expression for position. How do I go about calculating a numerical answer for these?

Thanks

Answer 1

NO the other person is wrong you integrate the V(t) vector to get the R(t)vector



then you put i-j as the LOCATION that is the location it starts and solve for the integration constants



dr = ((12t^3 + 2t)i - (4t)j + (6t^2 -4)k)dt



integrate



r = (3t^4 + t^2)i - (2t^2)j + (2t^3 - 4t)k



since we start at i-j



i - j = (3t^4 + t^2 + Xio)i -(2t^2 + Yjo)j + (2t^3 - 4t + Zko)k



since the z component is 0 start with that



0 = 2t^3 - 4t + Zko



time is relative so use t = 0

0 = 0 - 0 + Zko



there Zko = 0



-1 = -(2t^2 + Yjo)

at t = 0

-1 = -(0 + Yjo)

1 = Yjo



at t = 0

1 = (0 + 0 + Xio)



Xio = 1



therefore the equation for R vector is

R(t) = (3t^4 + t^2 + 1)i -(2t^2 + 1)j + (2t^3 - 4t )k



then just do the derivative for the A(t) vector



A(t) = (36t^2 + 2)i - 4j + 12tk



pretty easy

Answer 2

Substitute i-j for t and then do the calculations. After calculating it, you get the numerical values by substituting the values in this equation √[((x)i)^2 + ((y)j)^2 + ((z)k)^2].



For the velocity:

V(t) = (12t^3 + 2t)i - (4t)j + (6t^2 - 4)k

V(i-j) = (12(i-j)^3 + 2(i-j))i - (4(i-j))j + (6(i-j)^2 - 4)k

V=√[((x)i)^2 + ((y)j)^2 + ((z)k)^2].



Keep in mind that i*i=j*j=k*k=1 and i*j=j*k=i*k=0.



Have a nice day.

Answer 3

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