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
