IEMDaily - Video Lecture Notes

Formula

Constant acceleration in two dimensions

When the acceleration

\vec{a}

is constant we have two sets of equations to describe the x and y coordinates. In the following, motion of the particle begins at t = 0; the initial position of the particle is given by

{\vec{r}}_{o} = x_{o} \hat{i} + y_{o} \hat{j}

And its initial velocity is given by

{\vec{v}}_{o} = v_{o x} \hat{i} + v_{o y} \hat{j}

And vector

\vec{a} = a_{x} \hat{i} + a_{y} \hat{j}

is constant

Hence Equation becomes

v_{x} = v_{o x} + a_{x} t

x - x_{o} = v_{o x} + \frac{1}{2} a_{x} t^{2}

v_{x}^{2} = v_{o x}^{2} + 2 a_{x} (x - x_{o})

v_{y} = v_{o y} + a_{y} t

y - y_{o} = v_{o y} + \frac{1}{2} a_{y} t^{2}

v_{y}^{2} = v_{o y}^{2} + 2 a_{y} (y - y_{o})

Formula

Constant acceleration in three dimensions

When the acceleration

\vec{a}

is constant we have three sets of equations to describe the x, y and z coordinates. In the following, motion of the particle begins at t = 0; the initial position of the particle is given by

{\vec{r}}_{o} = x_{o} \hat{i} + y_{o} \hat{j} + z_{o} \hat{k}

And its initial velocity is given by

{\vec{v}}_{o} = v_{o x} \hat{i} + v_{o y} \hat{j} + v_{o z} \hat{k}

And vector

\vec{a} = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}

is constant

Hence Equation becomes

v_{x} = v_{o x} + a_{x} t

x - x_{o} = v_{o x} + \frac{1}{2} a_{x} t^{2}

v_{x}^{2} = v_{o x}^{2} + 2 a_{x} (x - x_{o})

v_{y} = v_{o y} + a_{y} t

y - y_{o} = v_{o y} + \frac{1}{2} a_{y} t^{2}

v_{y}^{2} = v_{o y}^{2} + 2 a_{y} (y - y_{o})

v_{z} = v_{o z} + a_{z} t

z - z_{o} = v_{o z} + \frac{1}{2} a_{z} t^{2}

v_{z}^{2} = v_{o z}^{2} + 2 a_{z} (z - z_{o})

Formula

Third equation of motion

Let,
s -displacement of object startig from rest
u - initial velocity
v - final velocity
t - time of travel
Acceleration is defined as the time rate of change of velocity.

a = \frac{d v}{d t}

multiply and divide numerator and denominator of R.H.S. by

d s

a = \frac{d v}{d t} \times \frac{d s}{d s}

\frac{d s}{d t} = v

hence

a = v \frac{d v}{d s}

a d s = v d v

integrating bothsides, we get

\int_{0}^{s} a d s = \int_{u}^{v} v d v

2 a s = v^{2} - u^{2}

v^{2} = u^{2} + 2 a s

......

(3^{r d} e q u a t i o n o f m o t i o n)

Formula

Distance Traveled in nth second

Distance traveled in

n^{t h}

second is given by:

S_{n} = u + 1 / 2 a (2 n - 1)

Example

Finding Stopping Distance

Let the initial velocity of the particle is

10 m / s

. It is retarding at a deceleration of

1 m / s^{2}

. Its final velocity is

0 m / s

. Then the stopping distance by this particle will be given by:

v^{2} = u^{2} + 2 a s

0^{2} = 10^{2} - 2 \times 1 \times s

s = 50 m

Stopping time will be:
from

v = u + a t

0 = 10 - 1 \times t

t = 10 s

Example

Solving problems involving more than one equation of motion

Let the initial velocity of a particle is

5 m / s

and it is accelerated for

2 s

with an acceleration of

1 m / s^{2}

. The distance covered by the particle during this motion can be estimated by:

v = u + a t

v = 5 + 1 \times 2^{2}

v = 9 m / s

Now

9^{2} = 5^{2} + 2 \times 1 \times s

s = 28 m

Example

Problems Involving First Equation of Motion

A body is moving with an initial velocity of

2 \hat{i} m / s

and acceleration of

- 0.5 \hat{i} m / s^{2}

. Find the time when the velocity of the particle becomes zero.
Final velocity,

v = 0

v = u + a t

0 = 2 \hat{i} - 0.5 \hat{i} t

t = 4 s

Definition

Problems on second equation of motion

If initial velocity of a particle is

2 m / s

and it travels

10 m

1 s

then what will be its acceleration?
Since

s = u t + 1 / 2 a t^{2}

10 = 2 \times 1 + 1 / 2 a \times 1^{2}

a = 16 m / s^{2}

Example

Problems on Third Equation of Motion

Let the initial velocity of the particle is

5 m / s

. It is accelerating at an acceleration of

1 m / s^{2}

. Its final velocity is

10 m / s

. Then the distance traveled by this particle will be given by:

v^{2} = u^{2} + 2 a s

10^{2} = 5^{2} + 2 \times 1 \times s

s = 37.5 m

Example

Use of first equation of motion

Example: Find the time taken by a particle in reaching a velocity of

5 m / s

moving at an acceleration of

2 m / s^{2}

with zero initial velocity.

Solution:

v = u + a t

5 = 0 + 2 \times t

t = 2.5 s

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