Homework SourseGiven that the acceleration vector is at9cos3ti9sin3tj3tk th
Given that the acceleration vector is a(t)=(-9cos(-3t))i+(-9sin(-3t))j+(3t)k, the initial velocity is v(0)=i+k, and the initial position vector is r(0)=i+j+k, compute:
The velocity vector v(t)=
The position vector r(t)=
Note: the coefficients in your answers must be entered in the form of expressions in the variable \\emph{t}; e.g. \"5 cos(2t)\"
The velocity vector v(t)=
The position vector r(t)=
Note: the coefficients in your answers must be entered in the form of expressions in the variable \\emph{t}; e.g. \"5 cos(2t)\"
Solution
a(t)= -9cos(3t) i + 9sin(3t) j+ (3t) k We integrate this w.r.t time to obtain the velocity vector. We get, v(t) = -3sin(3t) i - 3cos(3t) j + 3t^2/2 k + C In order to compute the integration constant C, we have a condition- v(0) = i + k So, we put t=0 in v(t) and equate it to v(0) 0 i - 3 j + 0 k + C = i + k C = i + 3j + k So, v(t) = -3sin(3t) i - 3cos(3t) j + 3t^2/2 k + i+3j+k v(t) = [-3sin3t + 1] i + [-3cos3t + 3] j + [3t^2/2 + 1] k -->Answer For position vector, we integrate velocity vector w.r.t time. We get, r(t) = [cos3t + t] i + [-sin3t + 3t] j + [1/2 t^3 + t] k + C\' Again, we have a condition- r(0) = i + j + k Putting t = 0 in r(t) and equating it to r(0), we get, i + 0j + 0k + C\' = i + j + k C\' = j + k So, r(t) = [cos3t + t+1] i + [-sin3t + 3t] j + [1/2 t^3 + t+1] k -->Answer
