You will have to remember certain basic formulae for solving multiple choice questions within the permitted time. The essential things you need to remember in

**are here:***one dimensional uniformly accelerated motion***(1)**The final velocity (

*v*) at time

*t*of an object in one dimensional motion with uniform acceleration

*a*is given by

**where**

*v*=*v*_{0}+*at**v*

_{0 }is the initial velocity (at time

*t*= 0)

[If you use the symbol

*u*for the initial velocity, the equation becomes**]***v*=*u*+*at***(2)**The position

*x*of the object at time

*t*is related its initial position

*x*

_{0}(at zero time) as

*x*=*x*_{0}+*v*_{0}*t*+ (½)*at*^{2}
This can be rewritten in terms of the displacement

*s*=*x*–*x*_{0}as**s**

*=**v*_{0}*t*+ (½)*at*^{2}
[If you use the symbol

*u*for the initial velocity, the above equation becomes**s***=**ut*+ (½)*at*^{2}]**(3)**The final velocity is related to the displacement

*(*

*x*–

*x*

_{0}) as

*v*^{2}=*v*_{0}^{2}+ 2*a*(*x***–**

*x*_{0})
[If you use the symbol

*u*for the initial velocity and the symbol*s*for the displacement, the above equation becomes**]***v*^{2}=*u*^{2}+ 2*a*s**(4)**The distance

*(*

*s*

_{n}= x_{n}–

*x*

_{n}

_{–}

_{1}) traveled during the

*n*second is given by

^{th}**=**

*s*_{n}**+**

*v*_{0}

*a*(*n***– ½)**

[If you use the symbol

*u*for the initial velocity, the above equation becomes**=***s*_{n}**+***u**a*(*n***– ½)**]**(5)**

**(i)**The slope of the displacement – time graph (obtained by plotting time on the X-axis and the displacement on the Y-axis) gives the velocity.

**(ii)**The slope of the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the acceleration.

**(iii)**The area under the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the displacement.

In

**two dimensional motion**you have to consider mainly circular motion and projectile motion. We have already discussed the essential points required in the case of circular motion in the post dated 20^{th}January 2008. You can access that post as well as related posts by clicking on the label ‘circular motion’ below this post. Here are the important equations you need to remember in the case of**:***projectile motion***(6)**In projectile motion, the gravitational force affects the vertical component of the velocity (of the projectile) only. If the projectile is launched with velocity

*v*

_{0}from the origin, making an angle

*θ*with respect to the horizontal, the x-component and the y-component of the velocity

_{0}*v*are respectively

_{0}*v*

_{0x}_{ }=

_{ }

*v*

_{0}_{ }cos

*θ*

_{0}and

*v*

_{0y}_{ }=

_{ }

*v*

_{0}_{ }sin

*θ*

_{0}

The x and y co-ordinates of the projectile after time

*t*are respectively*x = (v*

_{0}_{ }cos

*θ*

_{0})

*t*and

*y = (v*

_{0}_{ }sin

*θ*)

_{0}*t*– (½)

*gt*

^{2}

The x and y components of the velocity of the projectile after time

*t*are respectively*v*

_{x }=

_{ }

*v*

_{0}_{ }cos

*θ*and

_{0}*v*

_{y }=

_{ }

*v*

_{0}_{ }sin

*θ*

_{0}_{ }–

*gt*

**(7)**Time of flight(

*T*) of the projectile is given by

_{f}

*T*= (2_{f}*v*_{0}_{ }sin*θ*)_{0}_{ }**/**

*g***(8)**Horizontal range (

*R*) of the projectile is given by

*R =*(*v*_{0}^{2}sin 2*θ*)/_{0}*g***(i)**The horizontal range is maximum (

*R*) when the angle of projection

_{max}*θ*= 45º and

_{0}

*R*_{max}= v_{0}^{2}/^{ }*g*

**(ii)**If the velocity of projection

*v*is the same, the horizontal range is the same for angles of projection

_{0}*θ*and (90º –

_{0}*θ*). These directions are equally inclined to the 45º direction for maximum range.

_{0}**(10)**Maximum height (

*H*) reached by the projectile is given by

*H =*(

*v*

_{0}^{2}sin

^{2}

*θ*)/2

_{0}*g*

If the projectile is launched

*vertically*, the maximum height reached will be*v*_{0}^{2}/^{ }2*g*, which is*half**the maximum range*.**(11)**for any given value of the angle of projection

*θ*, the horizontal range

_{0}*R*and the maximum height

*H*are related as

*R =*4

*H*cot

*θ*

_{0}You can find useful posts on two dimensional motion here.

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