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The position of a particle is given by r=3.0thati + 2.0 t^(2)hatj+ 5.0 hatk . Where t is in seconds and the coefficients have the proper units for r to be in metres, (a) Find v(t) and aft) of the particle, (b) Find the magnitude and direction ofv(t) at t = 1.0 s. |
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Answer» Solution :`v(t)(d r)/(d t) = (d)/(t) (3.0 t hat(i) + 2.0 t^(2)hat(J) + 5.0 hat(K)) = 3.0 hat(i) + 4.0t hat(j)` `a(t) = (d v)/(d t) = +4.0 hat(j) a = 4.0 ms^(-2)` along y-direction (b) At `t = 1.0 s "" v = 3.0 hat(i) + 4.0 hat(j)` It.s magnitude is `v = SQRT(3^(2) + 4^(2)) + 5.0 ms^(-1)` and direction is `theta = tan^(-1)((v_(y))/(v_(x))) = tan^(-1)((4)/(3)) = 53^(@)` with x-axis. |
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