Aparticle moves in the xy plane with constant acceleration. at time zero, the particle is at x = 6.0 m, y = 10.0 m, and has velocity v = 1.0 m/s î + 6.0 m/s ĵ. the acceleration is given by the vector a = 5.0 m/s2 î + 7 m/s2 ĵ.
(a) find the velocity vector at t = 10.0 s.
(b) find the position vector at t = 1.0 s.
(c) give the magnitude and direction of the position vector in part (b).

Explanation:

Given that,

Position of the particle at t = 0,

Velocity of the particle at t = 0

Acceleration of the particle,

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

(c) Magnitude of y',

|y'| = 21.69 meters

Direction of the y',

Hence, this is the required solution.

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