# Motion in a plane with Constant Acceleration

## 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 earlier while studying motion in one dimension
Thus velocity is given by equation
$\vec{v}=\vec{v_0}+ \vec{a}t$                                           (8)
where
$\vec{v}$ is velocity vector
$\vec{v_0}$ is Initial velocity vector
$\vec{a}$ is Instantaneous acceleration vector

Similarly position is given by the equation
$\vec{r}- \vec{r_0}=\vec{v_0} t+ \frac {1}{2}\vec{a}t^2$                                           (9)
where $\vec{r_0}$ is Initial position vector
i,e
$r_0=x_0 \hat{i}+y_0 \hat{j}$
and average velocity is given by the equation
$\vec{v_{av}}=\frac {1}{2}(\vec{v}+ \vec{v_0}$                                           (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)
$x-x_0=v_{0x}t+ \frac {1}{2}a_xt^2$                                           (11b)
and
$v_y=v_{y0}+a_yt$                                           (12a)
$y-y_0=v_{0y} t + \frac {1}{2} a_y t^2$                                           (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 separate and simultaneous 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 $5.0 \hat{i}$ m/s and moves in x-yunder action of a force which roduces a constant acceleration of ($3.0 \hat{i} + 2.0 \vec{j}$) 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
$\vec{r}- \vec{r_0}=\vec{v_0} t+ \frac {1}{2}\vec{a}t^2$
Here $\vec{r_0}$ =0 ( As object starts from Origin)
$\vec{v_0}=5.0 \hat{i}$ m/s
$\vec{a}=3.0 \hat{i} + 2.0 \vec{j}$ m/s2
So
$\vec{r}=5.0 \hat{i} t + \frac {1}{2} (3.0 \hat{i} + 2.0 \vec{j})t^2$
$= (5t+1.5t^2) \hat {i}+ t^2 \hat{j}$
Now
$r=x \hat{i}+y \hat{j}$
Therefore,
$x(t)=5t+1.5t^2$ and $y(t)=t^2$
Given x=84
so $5t+1.5t^2 =84$
or t=6 sec
Then $y= t^2=36$ m
Now
$\vec{v}=\frac {d \vec{r}}{dt}=\frac {d}{dt}[(5t+1.5t^2) \hat{i} +t^2 \hat{j}] =(5+3t)\hat{i}+2t \hat{j}$
At t=6sec
$\vec{v}=23\hat{i} + 12 \hat{j}$
Speed =|v|=$\sqrt {(23^2+ 12^2) =26$ m/s