Find out nature of position vector:
The position vector of a specific at any instant t with respect to the axes x, y and z is specified by
r (t ) = (4 t^{ 2} + 5) i¯ + (3 t ^{2} + 2)j¯ + 0 k¯
Find out nature of its path and its position, velocity and acceleration at t = 3 seconds. Distances are measured in centimeters.
Solution
As the coefficient of k in the given equation is zero, there is no displacement along z direction at any time t. This may be described as a motion of a particle in plane (XOY). If there is any coefficient for the k vector of the motion, the motion of the particle shall be in space.
The given equation of motion can be written as
x (t ) = 4 t ^{2} + 5,
y (t) = 3 t^{ 2} + 2
Eliminating t^{2} from the above two relations, we obtain
t 2 = x - 5/4 = y - 2 / 3
∴ 3 x - 15 = 4 y - 8
∴ 3 x - 4 y - 7 = 0
The velocity and acceleration of the particle may be calculated from the relations
dx/ dt = d /dt (4 t^{ 2 }+ 5) = 8 t, dy / dt = d dt (3 t ^{2} + 2) = 6 t
d ^{2} x / dt ^{2} = 8, d ^{2 }y/ dt ^{2} = 6t
at t = 3 v = 24 i¯ + 18 j¯
= 30 cm / sec. (that means)
a = 8 i¯ + 6 j¯
= 10 cm / sec ^{2} .