# Motion in a plane with Constant Accleration

## 5. Motion in a plane with Constant Accleration

• 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
vav=(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
vx=vx0+axt                                          (11a)
x-x0=v0xt+(1/2)axt2                                          (11b)
and
vy=vy0+ayt                                           (12a)
y-y0=v0yt+(1/2)ayt2                                           (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 VA and VB.
Relative velocity is defined as
VBA=VB-VA
where VBA is relative velocity of B relative to A

Similiary relative velocity of A relative to B
VAB=VA-VB
we will need to add or subtracting components along x & y direction to get the relative velocity
Suppose
vA=vxai + vyaj
vB=vxbi + vybj

Relative velocity of B relative to A
=vxbi + vybj -(vxai + vyaj)
=i(vxb-vxa) + j(vyb-vya)

For three dimensions motion
vA=vxai + vyaj +vzaz
vB=vxbi + vybj + vzbz

Relative velocity of B relative to A
=vxbi + vybj + vzbz -(vxai + vyaj +vzaz)
=i(vxb-vxa) + j(vyb-vya)+z(vyb-vya)

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