MOTION IN A PLANE
Position Vector and Displacement
The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by
where x and y are
components of r along
x-, and y- axes or simply
they are the coordinates
of the object.
Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′. Then, the displacement is
and is directed from P to P′
We can write in a component form
Velocity
The average velocity of an object is the ratio of the displacement and the corresponding time interval
Since the direction of the average velocity is the same as that of ∆r
The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero
The direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.
The magnitude of v is then
and the direction of v is given by the angle θ
vx , vy and angle θ are shown in fig. for a velocity vector v at point p .
Acceleration
The average acceleration a of an object for a time interval ∆t moving in x-y plane is the change in velocity divided by the time interval
The acceleration (instantaneous acceleration) is the limiting value of the average acceleration as the time interval approaches zero
One dimension, the velocity and the acceleration of an object are always along the same straight line. However, for motion in two or three dimensions, velocity and acceleration vectors may have any angle between 0° and 180° between them.
Numerical
Q:The position of a particle is given by
where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0 s.
Ans:
MOTION IN A PLANE WITH CONSTANT ACCELERATION
Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an interval of time, the average acceleration will equal this constant value. Let the velocity of the object be v0 at time t = 0 and v at time t.
Let us now find how the position r changes with time. In the one dimensional case, let ro and r be the position vectors of the particle at time 0 and t and let the velocities at these instants be vo and v. Then, over this time interval t, the average velocity is(vo + v)/2.
The displacement is the average velocity multiplied by the time interval
It can be easily verified that the derivative of i.e. dr/dt gives above eq. and it also satisfies the condition that at t=0, r = ro .
Equation can be written in component form as
Motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.
Numerical
Q: A particle starts from origin at t = 0 with a velocity 5.0 î m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0 î +2.0 ĵ ) ) m/s2 . (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84 m ? (b) What is the speed of the particle at this time ?
Ans:r0 = 0, the position of the particle is given by