Question

A particle 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).

Answer 1

Explanation:

Given that,

Position of the particle at t = 0,

$y=(6i+10j)\ m$

Velocity of the particle at t = 0

$u=(1i+6j)\ m$

Acceleration of the particle,

$a=(5i+7j)\ m/s^2$

Solution,

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

$v=u+at$

$v=(1i+6j)+(5i+7j)10$

$v=(51i+76j)\ m/s$

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

$y'=y+ut+\dfrac{1}{2}at^2$

$y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2$

$y'=\dfrac{19}{2}i+\dfrac{39}{2}j$

(c) Magnitude of y',

$|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}$

|y'| = 21.69 meters

Direction of the y',

$tan\theta=\dfrac{y}{x}$

$tan\theta=\dfrac{39/2}{19/2}$

$\theta=64.02^{\circ}$

Hence, this is the required solution.