Motion in a plane

• 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=v₀+at              ----------------------(1)

where

v is velocity vector
v₀ is Initial velocity vector
a is Instantaneous acceleration vector

Similarly position is given by the equation

r-r₀=v₀t+(1/2)at²    --------------------(2)

where r₀ is Initial position vector i,e
r₀=x₀i+y₀j

Average velocity is given by the equation
v_avg=(1/2)(v+v₀)  ------------------------ (3)

Since we have assumed particle to be moving in x-y plane,the x and y components of equation (1) and (2) are

v_x=v_x0+a_xt                    ------------------------  (4)
x-x₀=v_0xt+(1/2)a_xt²       ------------------------  (5)

and

v_y=v_y0+a_yt                     ------------------------ (6)
y-y₀=v_0yt+(1/2)a_yt²       ------------------------  (7)

• from above equation 4,5 and 6,7 ,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)

1. A ball is projected from the origin. The x and y coordinates of its displacement are given by \(x=3t\) and \(y=4t-5{{t}^{2}}\). Find the velocity of projection (in m/s). (given \(x,~y\) in m and \(t\) in \(s\)  )

Solution: 

\({{v}_{x}}=\frac{dx}{dt}=3\)

\({{v}_{y}}=\frac{dy}{dt}=\,\,\,\,4-10t\)

\({{v}_{y}}\,\,\,\,\,ux=3,\,\,\,\,\,\,uy=4\)

\(U=~5~m/s.\)

2. In a projectile motion let \({{v}_{x}}\) and \({{v}_{y}}\) are the horizontal and vertical components of velocity at any time t and \(x\) and \(y\)  are displacements along horizontal and vertical from the point of projection at any time t. Then If \(u\) is the initial speed and \(\theta\)the angle of projection. Then

Solution: 
\({{v}_{y}}=u\) \(sin \theta -gt\)

\(i.e.,{{v}_{y}}-t\) graph is a straight line with negative slope and positive intercept.

\(x=v\cos \theta t\)

i.e.,\(x-t\) graph is a straight line passing through origin.

\(y-u\sin \theta t-\frac{1}{2}g{{t}^{2}}\)

i.e., \({{v}_{x}}-t\) graph is a straight line parallel to t-axis

3. A balloon rises from rest on the ground with constant acceleration \(\frac{g}{8}\) where ‘g’ is acceleration due to gravity.  A stone is dropped when the balloon has risen to a height of 39.2m.  Find the time taken by the stone to reach the ground in seconds.

Solution: 

\(h=\frac{1}{2}a{{t}^{2}}\Rightarrow t=\sqrt{\frac{2h}{a}}=\sqrt{\frac{2\times 39.2\times 8}{9.8}}=85\)

Velocity after 8 sec,\(V=\frac{g}{8}\times 8=gm{{s}^{-1}}\)

\(39.2=-gt+\frac{1}{2}g{{t}^{2}}\,\,\Rightarrow {{t}^{2}}-2t-8=0\,\,\,\Rightarrow t=4s\)

1. A bomb at rest at the summit of a cliff breaks into two equal fragments. One of the fragments attains a horizontal velocity of \(20\sqrt{3}m{{s}^{-1}}\) The horizontal distance between the two fragments when their  displacement vectors  is  inclined at \(60{}^\circ\)relative to  each other is \((g=10m{{s}^{-2}})\)

Solution: 

\(\begin{align} & {{{\bar{s}}}_{1}}={{u}_{1}}t\,\hat{i}-\frac{1}{2}g{{t}^{2}}\,\hat{j} \\ & {{{\bar{s}}}_{2}}=-{{u}_{2}}t\,\hat{i}-\frac{1}{2}g{{t}^{2}}\,\hat{j} \\ & \cos {{60}^{o}}=\frac{{{{\bar{s}}}_{1}}.{{{\bar{s}}}_{2}}}{{{s}_{1}}.{{s}_{2}}} \\ \end{align}\)

\(\begin{align} & {{s}_{x}}={{s}_{{{x}_{1}}}}+{{s}_{{{x}_{2}}}} \\ & {{s}_{x}}=({{u}_{1}}+{{u}_{2}})t \\ \end{align}\)

2. When a body is projected from a level ground the ratio of its speed in the vertical and horizontal direction is 4 : 3. If the velocity of projection is u, the time after which, the ratio of the velocities in the vertical and horizontal directions are reversed is

Solution: 

\({{u}_{x}}:{{u}_{y}}=\,\,\,3:4\)

\({{u}_{x}}=\frac{3u}{5}\)   ; \({{v}_{x}}=\frac{3u}{5}\)

\({{u}_{y}}=\frac{4u}{5}\)   ; \({{v}_{y}}=\frac{3}{4}{{v}_{x}}=\frac{9}{20}u\)

\(\frac{9}{20}u=\frac{4u}{5}-gt\)

gt =\(\frac{16u}{20}-\frac{9u}{20}\)   \(\Rightarrow \frac{7u}{20}\)

t =\(\frac{7u}{20g}\)

3. A body is thrown with velocity (4i + 3j) metre per second. Its maximum height \(g=10m{{s}^{-2}}\)

Solution: 

\({{h}_{\max }}=\frac{v_{y}^{2}}{2g}=\frac{{{(3)}^{2}}}{20}=\frac{9}{20}=0.45m\)  

1. The horizontal  range and maximum height attained by a projectile are R and H respectively. If a constant horizontal  acceleration \(a=\frac{g}{2}\) is imparted to the projectile  due to wind then its horizontal  range and maximum  height will be

Solution:

\(H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{{{u}^{2}}_{y}}{2g}\)

Since\({{u}_{y}} \,\,and \,\,{{a}_{y}}\)remain  the same as previous  so H remains  same

So T also  remains  the same

Now \({{R}_{new}}={{u}_{x}}T+\frac{1}{2}{{a}_{x}}{{T}^{2}}\)

\(\Rightarrow {{R}_{new}}=R+\frac{1}{2}\left( \frac{g}{2} \right)\left( \frac{4{{u}^{2}}}{{{g}^{2}}} \right)=R+2H\)

2. A particle  is moving  in x-y plane. At certain instant  of time the components of  its velocity  and acceleration are \({{v}_{x}}=3m{{s}^{-1}}, {{v}_{y}}=4m{{s}^{-1}}, {{a}_{x}}=2m{{s}^{-2}}\) and \({{a}_{y}}=1m{{s}^{-2}}\) ,The  rate of change of speed at this moment is

Solution: 

Since \({{v}^{2}}={{v}_{x}}^{2}+{{v}_{y}}^{2}\)

\(\Rightarrow 2v\frac{dv}{dt}=2{{v}_{x}}\frac{d{{v}_{x}}}{dt}+2{{v}_{y}}\frac{d{{v}_{y}}}{dt}\)

\(\Rightarrow \frac{dv}{dt}=\frac{{{v}_{x}}{{a}_{x}}+{{v}_{y}}{{a}_{y}}}{v}=\frac{{{v}_{x}}{{a}_{x}}+{{v}_{y}}{{a}_{y}}}{\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}}\)

\(\Rightarrow \frac{dv}{dt}=\frac{\left( 3 \right)\left( 2 \right)+\left( 4 \right)\left( 1 \right)}{5}=2m{{s}^ {-1}}\)