5. Motion in a plane with Constant Acceleration

5. Motion in a plane with Constant Acceleration

• Motion in two dimension with constant acceleration we we know is the motion in which velocity changes at a constant rate i.e, acceleration remains constant throughout the motion
• We should set up the kinematic equation of motion for particle moving with constant acceleration in two dimensions.
• Equation's for position and velocity vector can be found generalizing the equation for position and velocity derived earliar while studying motion in one dimension
  Thus velocity is given by equation
  v=v₀+at                                          (8)
  where
  v is velocity vector
  v₀ is Intial velocity vector
  a is Instantanous acceleration vector
  Similary position is given by the equation
  r-r₀=v₀t+(1/2)at²                                          (9)
  where r₀ is Intial position vector
  i,e
  r₀=x₀i+y₀j
  and average velocity is given by the equation
  v_av=(1/2)(v+v₀)                                          (10)
• Since we have assumed particle to be moving in x-y plane,the x and y components of equation (8) and (9) are
  v_x=v_x0+a_xt                                          (11a)
  x-x₀=v_0xt+(1/2)a_xt²                                          (11b) 
  and 
  v_y=v_y0+a_yt                                           (12a)
  y-y₀=v_0yt+(1/2)a_yt²                                           (12b)
• from above equation 11 and 12 ,we can see that for particle moving in (x-y) plane although plane of motion can be treated as two seperate and simultanous 1-D motion with constant acceleration
• Similar result also hold true for motion in a three dimension plane (x-y-z)

Question
A object starts from origin at t = 0 with a velocity 5.0 i m/s and moves in x-yunder action of a force which roduces a constant acceleration of (3.0i + 2.0j) 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 ?
Solution
We know that position of the object is given by
r-r₀=v₀t+(1/2)at²
Here r₀ =0 ( As object starts from Origin)
v₀=5.0 i m/s
a=(3.0i + 2.0j) m/s²
So
r=5.0 i t+(1/2)(3.0i + 2.0j)t²
= (5t+1.5t²)i+t²j
Now
r =x(t)i + y(t)j
Therefore,
x(t)=5t+1.5t² and y(t)=t²
Given x=84
so 5t+1.5t² =84
or t=6 sec
Then y= t²=36 m
Now
v=dr/dt=d[(5t+1.5t²)i+t²j]/dt =(5+3t)i+2tj
At t=6sec
v=23i+ 12j
Speed =|v|=√(23²+ 12²) =26m/s

Relative velocity in Two Dimension

For two objects A and B moving with the uniform velocities V_A and V_B.
Relative velocity is defined as
V_BA=V_B-V_A
where V_BA is relative velocity of B relative to A

Similiary relative velocity of A relative to B
V_AB=V_A-V_B
we will need to add or subtracting components along x & y direction to get the relative velocity
Suppose
v_A=v_xai + v_yaj
v_B=v_xbi + v_ybj

Relative velocity of B relative to A
=v_xbi + v_ybj -(v_xai + v_yaj)
=i(v_xb-v_xa) + j(v_yb-v_ya)

For three dimensions motion
v_A=v_xai + v_yaj +v_zaz
v_B=v_xbi + v_ybj + v_zbz

Relative velocity of B relative to A
=v_xbi + v_ybj + v_zbz -(v_xai + v_yaj +v_zaz)
=i(v_xb-v_xa) + j(v_yb-v_ya)+z(v_yb-v_ya)