## 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 earlier while studying motion in one dimension

Thus velocity is given by equation

**v**=**v**_{0}+**a**t (8)

where

**v** is velocity vector

**v**_{0} is Initial velocity vector

**a** is Instantaneous acceleration vector
Similarly position is given by the equation

**r**-**r**_{0}=**v**_{0}t+(1/2)**a**t^{2} (9)

where **r**_{0} is Initial position vector

i,e

**r**_{0}=x_{0}**i**+y_{0}**j**

and average velocity is given by the equation

**v**_{av}=(1/2)(**v**+**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_{x}t (11a)

x-x_{0}=v_{0x}t+(1/2)a_{x}t^{2} (11b)

and

v_{y}=v_{y0}+a_{y}t (12a)

y-y_{0}=v_{0y}t+(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 **i** m/s and moves in x-yunder action of a force which roduces a constant acceleration of (3.0**i** + 2.0**j**) m/s^{2}

(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**-**r**_{0}=**v**_{0}t+(1/2)**a**t^{2}

Here **r**_{0} =0 ( As object starts from Origin)

**v**_{0}=5.0 **i** m/s

**a**=(3.0**i** + 2.0**j**) m/s^{2}

So

**r**=5.0 **i** t+(1/2)(3.0**i** + 2.0**j**)t^{2}

= (5t+1.5t^{2})**i**+t^{2}**j**

Now

**r** =x(t)**i** + y(t)**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

**v**=d**r**/dt=d[(5t+1.5t^{2})**i**+t^{2}**j**]/dt =(5+3t)**i**+2t**j**

At t=6sec

**v**=23**i**+ 12**j**

Speed =|**v**|=√(23^{2}+ 12^{2}) =26m/s

**link to this page by copying the following text**
**Also Read**

### Search Our Website