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Chapter-2
Motion along a straight line
Prepared by Dr
...
K
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It is a scalar quantity and cannot be
negative
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Displacement Vector
The shortest distance between the initial and final point is called displacement
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If during a time interval ∆t the position vector of the particle changes from ‫ݔ‬ଵ to ‫ݔ‬ଶ, the
Ԧ
Ԧ
Displacement vector of the particle ߂‫ ݔ‬for that time interval is defined as: ߂‫ݔ = ݔ‬ଶ − ‫ݔ‬ଵ
Ԧ
Ԧ
Ԧ
Ԧ
Speed
The amount of distance travelled by a body in one second is called its speed
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It cannot be negative
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ܽ‫= ݀݁݁݌ݏ ݁݃ܽݎ݁ݒ‬

‫ݔ ݈݈݀݁݁ݒܽݎݐ ݁ܿ݊ܽݐݏ݅݀ ݈ܽݐ݋ݐ‬ଵ + ‫ݔ‬ଶ + ‫ݔ‬ଷ …
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Problem
A body moves with speed 50 km/h and returned with 40 km/h to the same point
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ܽ‫= ݀݁݁݌ݏ ݁݃ܽݎ݁ݒ‬

‫ݔ‬ଵ + ‫ݔ‬ଶ
‫ݔ+ݔ‬
= ‫ݔ‬
‫/݉݇ 4
...


ܽ‫= ݕݐ݅ܿ݋݈݁ݒ ݁݃ܽݎ݁ݒ‬
‫ݒ‬௔௩௚ି௫ =
Ԧ

Ԧ
௱௫
௱௧

=

ሬሬሬሬԦି௫భ
௫మ ሬሬሬሬԦ
௧మ ି௧భ

ܿℎܽ݊݃݁ ݅݊ ݀݅‫ݐ݈݊݁݉݁ܿܽ݌ݏ‬
‫݈ܽݒݎ݁ݐ݊݅ ݁݉݅ݐ‬

Working formula

Instantaneous Velocity:
As discussed in rectilinear motion, a more interesting quantity is the instantaneous velocity ‫ݒ‬
Ԧ,
which is the limit of the average velocity when we shrink the time interval ∆t to zero
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The velocity ‫ݒ‬ଶ differs both in magnitude and direction from the velocity ‫ݒ‬ଵ
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Ԧ
Ԧ
Instantaneous Acceleration

But the much more interesting quantity is the result of shrinking the period ߂‫ ݐ‬to zero, which
gives us the instantaneous acceleration, ܽ It is the time derivative of the velocity vector ‫ݒ‬
Ԧ
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We want to derive two/three sets of equations to describe the x, y and z coordinates, each of which is similar to the equations in rectilinear motion
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௩ ି௩
Therefore, the acceleration along x-direction will be, ܽ௫ = మೣ ି௧ భೣ
...

Then, ܽ௫ =

௩ೣ ି௩బೣ
௧ି଴

‫ݒ‬௫ = ‫ݒ‬଴௫ + ܽ௫ ‫ )1
...
Then the average velocity will be
‫ݒ‬௔௩ି௫ =

‫ݔ − ݔ‬଴
… … … … … … … …
...
2)
‫ݐ‬

If the same body moves with initial velocity ‫ݒ‬଴௫ and attain final velocity ‫ݒ‬௫ after certain time,
then average velocity will be
‫ݒ‬௔௩ି௫ =
‫ݒ‬௔௩ି௫ =

௩బೣ ା௩ೣ
ଶ

(‫ݒ‬଴௫ + ‫ݒ‬଴௫ + ܽ௫ ‫)ݐ‬
2

1
⇒ ‫ݒ‬௔௩ି௫ = ‫ݒ‬଴௫ + ܽ௫ ‫)3
...
… … … … … … … ݐ‬
2
From eq
...
2) and eq
...
3), we get

‫ݔ − ݔ‬଴
1
= ‫ݒ‬଴௫ + ܽ௫ ‫ ݐ‬
‫ݐ‬
2

1
⇒ ‫ݔ − ݔ‬଴ = ‫ ݐ‬൬‫ݒ‬଴௫ + ܽ௫ ‫ݐ‬൰
2
ଵ

‫ݔ − ݔ‬଴ = ‫ݒ‬଴௫ ‫ܽ + ݐ‬௫ ‫ ݐ‬ଶ … … … … … … … … … …
...
4) Working formula
ଶ

Derivation of 3rd kinematic equation:
To relate displacement, acceleration, initial velocity and final velocity, we should find the time
from equation (1
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(1
...
(1
...
6) Working formula
The above eq
...
1),(1
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6) can be resolved in to three sets of equations to describe the
motion along the three Cartesian directions as
‫ݒ‬௫ = ‫ݒ‬଴௫ + ܽ௫ ‫ݒ ;ݐ‬௬ = ‫ݒ‬଴௬ + ܽ௬ ‫ݒ ;ݐ‬௭ = ‫ݒ‬଴௭ + ܽ௭ ‫ ;ݐ‬
1
1
1
(‫ݔ − ݔ‬଴ ) = ‫ݒ‬଴௫ ‫ܽ + ݐ‬௫ ‫ ݐ‬ଶ ; (‫ݕ − ݕ‬଴ ) = ‫ݒ‬଴௬ ‫ܽ + ݐ‬௬ ‫ ݐ‬ଶ ; (‫ݖ − ݖ‬଴ ) = ‫ݒ‬଴௭ ‫ܽ + ݐ‬௭ ‫ ݐ‬ଶ ;
2
2
2
‫ݒ‬௫ ଶ = ‫ݒ‬଴௫ ଶ + 2ܽ௫ (‫ݔ − ݔ‬଴ ); ‫ݒ‬௬ ଶ = ‫ݒ‬଴௬ ଶ + 2ܽ௬ (‫ݕ − ݕ‬଴ ); ‫ݒ‬௭ ଶ = ‫ݒ‬଴௭ ଶ + 2ܽ௭ (‫ݖ − ݖ‬଴ );

Best of luck



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