UNIT 1: Vectors

Physical quantities can be classified as either a scalar or a vector. Scalar quantities are quantizing
which are completely specified by their magnitude and units (of measurement); no direction is
required.
Examples: Distance(𝑆), speed(𝑣), mass(𝑀), Pressure(𝑃), electric current(𝐼), Voltage(𝑉) …

A vector is a quantity that is specified by both a magnitude and direction in space. Vectors obey
the laws of vector algebra.
Examples: displacement, velocity, acceleration, momentum, etc.

Representation of vector
A. Algebraic Method
Vectors are represented algebraically by a letter (or symbol) with an arrow over its head
(Example: velocity by𝑣⃗, momentum by𝑝⃗) and the magnitude of a vector is a positive scalar
and is written as either by |A| or A.
When dealing with vectors it is often useful to draw with an arrow.
B. Geometric Method

✓ Vectors are nothing but straight arrows drawn from one point to another.
✓ Zero vectors are just a vector of zero length - a point.
✓ The longer the arrow the bigger the magnitude, that vectors can be parallel transported around.

Vector Addition

A single vector that is obtained by adding two or more vectors is called resultant vector and it is
obtained using the following two methods.

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 A. Graphical method of vector addition
Added by joining their head to tail and in any order their resultant vector is the vector drawn from
the tail of the first vector to the head of the last vector.

B. Parallelogram law of vector addition

It states that the resultant R of two vectors A and B is the diagonal of the parallelogram for which
the two vectors A and B becomes adjacent sides as shown in figure below.

The magnitude of the resultant vector is obtained using cosine law and direction using the sine
law, respectively as follows.

When two/more vectors are added, the sum is independent of the order of the addition
(commutative law of addition). 𝐴 + 𝐵 = 𝐵 + 𝐴 = 𝑅

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When three or more vectors are added, their sum is independent of the way in which the individual
vectors are grouped (associative law of addition). +𝐵 + 𝐶 = 𝐴 + (𝐵 + 𝐶)

Components of Vector

Let us consider components of vector A is obtained by applying the trigonometric functions of

sine and cosine.

The components Ax and Ay can be added to give back A as their resultant.

Because 𝐴𝑥 and 𝐴𝑦 are perpendicular to each other, the magnitude of their resultant vector is
obtained using Pythagoras theorem.

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Similarly, any three-dimensional vector 𝐴 can be written as the sum of its 𝑥, 𝑦 ∧ 𝑧 components.
𝐴 = 𝐴𝑥 + 𝐴𝑦 + 𝐴𝑧
And its magnitude becomes

The direction angles that this vector makes with the three axes, is given by the direction of cosines.

Unit Vector

A unit vector is a vector that has magnitude of one and it is dimensionless.

It is usually denoted with a “hat”.

The unit vectors are so prevalent that we give them special names. For a two-dimensional 𝑥 − 𝑦
coordinate system we have the unit vector ^𝑖 pointing in the +𝑥 direction, and the unit vector ^𝑗 pointing
in the +𝑦 direction. For a three-dimensional 𝑥 −, 𝑦 − and 𝑧 − coordinate system, we have those two,
and one more, namely the unit vector 𝑘^ pointing in the +𝑧 direction.
Any vector can be expressed in terms of unit vectors. Consider, for instance, a vector A with
components𝐴𝑥 , 𝐴𝑦 , and 𝐴𝑧 . The vector formed by the product 𝐴𝑥 has magnitude |𝐴𝑥 | in the +x
direction. This means that 𝐴𝑥^𝑖 is the 𝑥 −component of vector A. Similarly, 𝐴𝑦^𝑗 is the 𝑦 −component
of vector A and, 𝐴𝑧 𝑘^ is the 𝑧 −component vector of A. Thus, A can be expressed as:

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Finding a Unit Vector

Consider the vector𝑟⃗ = 𝑥𝑖^ + 𝑦𝑗^ + 𝑧𝑘^. The unit vector 𝑟^ in the same direction as the vector is simply
the vector 𝑟⃗ divided by its magnitude r.
^ 𝑥
𝑟⃗ 𝑥𝑖^ + 𝑦𝑗^ + 𝑧𝑘 𝑦 𝑧
𝑟^ = = = ^𝑖 + ^𝑗+ 𝑘^
𝑟 𝑟 𝑟 𝑟 𝑟
C. Vector addition in Unit Vector Notation

Adding vectors that are expressed in unit vector notation is easy in that individual unit vectors
appearing in each of two or more terms can be factored out.

Negative of a vector

The negative of the vector 𝐴 is defined as the vector that when added to 𝐴 gives zero for the vector
sum. The negative of a given vector is a vector that has the same magnitude as that vector but
directed in opposite direction. 𝐴 + (−𝐴) = 0. The angle between the two vectors is180°.
Subtraction of vectors

✓ The subtraction vectors are an operation involving vector addition.

✓ 𝐴 − 𝐵 is the same as adding vector 𝐴 to the negative of vector 𝐵.
i.e., 𝐴 − 𝐵 = 𝐴 + (−𝐵)

Vector multiplication

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Vectors can be multiplied in two different ways i.e., dot product and cross product. The results in
both multiplications of vectors are different.

➢ Dot (Scalar) product

The dot product of A and B is:

𝐴⃗. 𝐵
⃗⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 = 𝐴𝐵𝑐𝑜𝑠𝜃 = 𝐴𝐵𝑐𝑜𝑠(𝛼 + 𝛽)
𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 𝐴𝑥 𝐵𝑥 𝐴𝑦 𝐵𝑦 𝐴𝑥 𝐵𝑥 𝐴𝑦 𝐵𝑦
𝑐𝑜𝑠(𝛼 + 𝛽) = = + = ( ) ( ) + ( ) ( ) = 𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 − 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽
𝐴𝐵 𝐴𝐵 𝐴𝐵 𝐴 𝐵 𝐴 𝐵

➢ Cross (Vector) product

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The cross product of A and B is:
𝐴⃗ × 𝐵 ^ = 𝐴𝐵𝑠𝑖𝑛𝜃 = 𝐴𝐵𝑠𝑖𝑛(𝛼 + 𝛽)𝑘
⃗⃗ = (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 )𝑘 ^
𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 𝐴𝑥 𝐵𝑦 𝐴𝑦 𝐵𝑥 𝐴𝑥 𝐵𝑦 𝐴𝑦 𝐵𝑥
𝑠𝑖𝑛(𝛼 + 𝛽) = = − = ( ) ( ) − ( ) ( ) = 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 − (−𝑠𝑖𝑛𝛼)𝑐𝑜𝑠𝛽
𝐴𝐵 𝐴𝐵 𝐴𝐵 𝐴 𝐵 𝐴 𝐵
= 𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 + 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽

➢ Triple scalar product

➢ Triple vector product

Example 1: Vector 𝐴⃗ has magnitude of 3𝑢𝑛𝑖𝑡𝑠 and makes an angle of 45° with the positive x-axis.
⃗⃗ has the magnitude of 4𝑢𝑛𝑖𝑡𝑠 and makes an angle of 45° with the negative x-axis. Find:
Vector 𝐵
a) The resolutions and magnitude of 𝐴⃗ and 𝐵
⃗⃗

b) The magnitude and direction of 𝐴⃗ + 𝐵

⃗⃗

c) The magnitude and direction of 𝐴⃗ − 𝐵

⃗⃗

d) The dot and cross products of 𝐴⃗ and 𝐵

⃗⃗

Solution:
a) The resolution and magnitude of 𝐴⃗ and 𝐵
⃗⃗

𝐴⃗ = 𝐴(𝑐𝑜𝑠(45°)𝑖^ + 𝑠𝑖𝑛(45°)𝑗^) = 2.121(𝑖^ + ^

𝑗)|𝐴⃗| = 2.121√2
⃗⃗ = 𝐵(−𝑐𝑜𝑠(45°)𝑖^ + 𝑠𝑖𝑛(45°)𝑗^) = −2.828(𝑖^ − ^
𝐵 ⃗⃗| = 2.828√2
𝑗)|𝐵

b) The magnitude of 𝑅⃗⃗1 = 𝐴⃗ + 𝐵

⃗⃗:

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 𝑅⃗⃗1 = 𝐴⃗ + 𝐵
⃗⃗ = 2.121(𝑖^ + ^
𝑗) + 2.828(−𝑖^ + ^
𝑗) = −0.707𝑖^ + 4.949𝑗^

|𝑅⃗⃗1 | = √(−0.707)2 + (4.949)2 = 4.999

c) The magnitude of 𝑅⃗⃗2 = 𝐴⃗ − 𝐵

⃗⃗:

𝑅⃗⃗2 = 𝐴⃗ − 𝐵
⃗⃗ = 2.121(𝑖^ + ^
𝑗) − 2.828(−𝑖^ + ^
𝑗) = 4.949𝑖^ − 0.707𝑗^

|𝑅⃗⃗2 | = √(4.949)2 + (−0.707)2 = 4.999

d) The dot and cross products of 𝐴⃗ and 𝐵

⃗⃗

The dot product of A and B is:

𝐴⃗. 𝐵
⃗⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 = (2.121)(−2.828) + (2.121)(2.828) = 0

The cross product of A and B is:

𝐴⃗ × 𝐵 ^ = {(2.121)(2.828) − (2.121)(−2.828)}𝑘
⃗⃗ = (𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 )𝑘 ^ = 11.996𝑘
^

Example 2: Given the displacement vectors𝐴⃗ = 3𝑖^ − 4𝑗^ + 4𝑘^ and𝐵 ^. Find the
⃗⃗ = 2𝑖^ + 3𝑗^ − 7𝑘

magnitudes of the vectors:

a) 𝐴⃗ + 𝐵
⃗⃗

b) 2𝐴⃗ − 𝐵
⃗⃗

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UNIT 2: 1- and 2-Danmessional Motions
A formal study of physics begins with kinematics. The word “kinematics” comes from a Greek
word “kinesis” meaning motion. Kinematics is concerned on analyzing kinematical quantities used
to describe motion such as velocity, acceleration, displacement, time, and trajectory. Objects are
in motion all around us.

Displacement, velocity and Acceleration in 1D and 2D

Position: - The location of an object with respect to a chosen reference point.

Displacement: - The change in position of an object with respect to a given reference frame.

Distance (S):- The length of the path followed by the object.

Average and Instantaneous Velocities:

Average Velocity ( 𝑣⃗𝑎𝑣 ):- is the change of displacement divided by the change of time.

∆𝑟⃗ 𝑟⃗𝑓 − 𝑟⃗𝑖

𝑣⃗𝑎𝑣 = =
∆𝑡 𝑡𝑓 − 𝑡𝑖

Average Speed: - is the total distance traveled by the object divided by the total elapsed time.

𝑡𝑜𝑡𝑎𝑙𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒(𝑆)
𝑣𝑎𝑣 =
𝑡𝑜𝑡𝑎𝑙𝑡𝑖𝑚𝑒𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙(∆𝑡)

Average speed and average velocity of an object do not provide the detail information of the entire
motion. We may need to know the velocity or speed of the particle at a certain instant of time.

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 ∆𝑟⃗
Instantaneous Velocity (𝑣⃗𝑖𝑛𝑠𝑡 ): - is the limiting value of the ratio ∆𝑡 as ∆𝑡 approaches zero

∆𝑟⃗ 𝑑𝑟
𝑣⃗ = 𝑙𝑖𝑚 =
∆𝑡→0 ∆𝑡 𝑑𝑡

Instantaneous speed: - It is the magnitude of the instantaneous velocity.

Average and Instantaneous Accelerations:

If the velocity of a particle changes with time, then the particle is said to be accelerating.

Average acceleration: is the change in velocity ( ∆𝑣⃗) of an object divided by the time interval
during which that change occurs.

∆𝑣⃗ 𝑣⃗𝑓 − 𝑣⃗𝑖

𝑎⃗𝑎𝑣 = =
∆𝑡 𝑡𝑓 − 𝑡𝑖

Instantaneous acceleration: -The limit of average acceleration as∆𝑡 approaches zero.

∆𝑣⃗ 𝑑𝑣
𝑎⃗ = 𝑙𝑖𝑚 =
∆𝑡→0 ∆𝑡 𝑑𝑡

Example 1: A particle moves according to the equation of 𝑟(𝑡) = (5𝑡 2 + 2𝑡)𝑚. Find:

a) The velocity and acceleration of a particle for time 𝑡.

b) The averages velocity and acceleration of the particle for time interval 1𝑠 to3𝑠.

Solution

∆𝑟⃗ 𝑑(5𝑡 2 +2𝑡)𝑚

a. 𝑣⃗(𝑡) = 𝑙𝑖𝑚 = = (10𝑡 + 2) 𝑚⁄𝑠
∆𝑡→0 ∆𝑡 𝑑𝑡
⃗⃗
∆𝑣 𝑑𝑣 𝑑(10𝑡+2)𝑚⁄𝑠
𝑎⃗(𝑡) = 𝑙𝑖𝑚 = = = 10 𝑚⁄𝑠 2
∆𝑡→0 ∆𝑡 𝑑𝑡 𝑑𝑡

b. 𝑟(𝑡) = (5𝑡 2 + 2𝑡)𝑚, → 𝑎𝑡𝑡 = 1𝑠, 𝑟(1) = 7𝑚, → 𝑎𝑡𝑡 = 3𝑠, 𝑟(3) = 51𝑚
∆𝑟⃗ 𝑟⃗𝑓 −𝑟⃗𝑖 51𝑚−7𝑚
𝑣⃗𝑎𝑣 = = = = 22 𝑚⁄𝑠
∆𝑡 𝑡𝑓 −𝑡𝑖 3𝑠−1𝑠

⃗⃗
∆𝑣 ⃗⃗𝑓 −𝑣
𝑣 ⃗⃗𝑖
𝑎⃗𝑎𝑣 = ∆𝑡
= 𝑡𝑓 −𝑡𝑖
= 10 𝑚⁄𝑠 2

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Example 2: A particle moves by a constant speed of 12 𝑚⁄𝑠 along the straight line from point A
to point B, and then back along the same line from B to A at a constant speed of 9 𝑚⁄𝑠. Find the
average speed and velocity of the particle over total movement of the particle.
Solution

𝑆𝑡𝑜𝑡 = 𝑆1 + 𝑆2 = 2𝑥, 𝑡𝑡𝑜𝑡 = 𝑡1 + 𝑡2 = 𝑡𝐴𝐵 + 𝑡𝐵𝐴

𝑆1 𝑥 𝑆2 𝑥
→ 𝑡𝐴𝐵 = = ∧ 𝑡𝐵𝐴 = =
⁄
𝑣1 12 𝑚 𝑠 𝑣2 9 𝑚⁄𝑠

𝑆𝑡𝑜𝑡 2𝑥 2𝑥
𝑣𝑎𝑣 = =
𝑡𝑡𝑜𝑡 𝑡𝐴𝐵 + 𝑡𝐵𝐴
= 𝑥 𝑥 = 10.3 𝑚⁄𝑠
+
12 𝑚⁄𝑠 9 𝑚⁄𝑠

∆𝑥⃗
𝑣⃗𝑎𝑣 = = 0, 𝑠𝑖𝑛𝑐𝑒𝑥𝑖 = 𝑥𝑓 = 𝑥
∆𝑡

Motion with Constant Acceleration

For motion with constant acceleration,

✓ The velocity changes at the same rate throughout the motion.

✓ Average acceleration over any time interval is equal to the instantaneous acceleration at
any instant of time.
𝑑𝑣 ∆𝑣⃗ 𝑣⃗𝑓 − 𝑣⃗𝑖
𝑎⃗ = = = , 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔𝑡𝑖 = 0
𝑑𝑡 ∆𝑡 𝑡
Rearranging this equation gives
𝑣⃗𝑓 = 𝑣⃗𝑖 + 𝑎⃗𝑡
For motion with constant acceleration, average velocity can be written as:

𝑣⃗𝑓 + 𝑣⃗𝑖 𝑑𝑟 ∆𝑟⃗

𝑣⃗𝑎𝑣 = , 𝑣⃗𝑎𝑣 = = , 𝑓𝑜𝑟𝑡𝑖 = 0 → ∆𝑟⃗ = 𝑣⃗𝑎𝑣 𝑡
2 𝑑𝑡 ∆𝑡

(𝑣⃗𝑓 + 𝑣⃗𝑖 )𝑡
→ 𝑟⃗𝑓 − 𝑟⃗𝑖 =
2

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 1
𝑟⃗𝑓 − 𝑟⃗𝑖 = 𝑣𝑖 𝑡 + 𝑎⃗𝑡 2
2

⃗⃗𝑓 +𝑣
𝑣 ⃗⃗𝑖 ⃗⃗𝑓 −𝑣
𝑣 ⃗⃗𝑖 ⃗⃗𝑓 +𝑣
(𝑣 ⃗⃗𝑖 ) (𝑣
⃗⃗𝑓 −𝑣
⃗⃗𝑖 )
Again, ∆𝑟⃗ = 𝑣⃗𝑎𝑣 𝑡𝑏𝑢𝑡𝑣⃗𝑎𝑣 = 2
∧𝑡= 𝑎
, ∆𝑟⃗ = 2 𝑎

𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑟

For 2D motion 𝑎⃗ = 𝑎𝑥^𝑖 + 𝑎𝑦^𝑗, 𝑣⃗𝑓 = 𝑣𝑥𝑓^𝑖 + 𝑣𝑦𝑓^𝑗, 𝑣⃗𝑖 = 𝑣𝑥𝑖^𝑖 + 𝑣𝑦𝑖^𝑗
𝑣⃗𝑥𝑓 = 𝑣⃗𝑥𝑖 + 𝑎⃗𝑥 𝑡
𝑣⃗𝑓 = 𝑣⃗𝑖 + 𝑎⃗𝑡 → { }
𝑣⃗𝑦𝑓 = 𝑣⃗𝑦𝑖 + 𝑎⃗𝑦 𝑡

Example 1: At𝑡 = 0𝑠, a particle moving in the x-y plane with constant acceleration has a velocity
of 𝑣⃗𝑖 = (3𝑖^ − 2𝑗^) 𝑚⁄𝑠 and is at the origin. At𝑡 = 3𝑠, the particle‘s velocity is𝑣⃗𝑓 = (9𝑖^ + 7𝑗^) 𝑚⁄𝑠.
Find:
a) the acceleration of the particle
b) Its coordinates at𝑡 = 3𝑠.
Solution:
⃗⃗
∆𝑣 ⃗⃗𝑓 −𝑣
𝑣 ⃗⃗𝑖 ((9𝑖^+7𝑗^)𝑚⁄𝑠)−((3𝑖^−2𝑗^)𝑚⁄𝑠)
a) 𝑎⃗ = ∆𝑡 = 𝑡
= 3𝑠
= (2𝑖^ + 3𝑗^) 𝑚⁄𝑠 2
1 1
b) 𝑟⃗𝑓 = 𝑟⃗𝑓 − 𝑟⃗𝑖 = 𝑣𝑖 𝑡 + 2 𝑎⃗𝑡 2 = (3𝑖^ − 2𝑗^) 𝑚⁄𝑠 (3𝑠) + 2 (2𝑖^ + 3𝑗^) 𝑚⁄𝑠 2 (3𝑠)2
𝑟⃗𝑓 = (18𝑖^ − 7.5𝑗^)𝑚
So, at𝑡 = 3𝑠 (𝑥, 𝑦) = (18, −7.5)𝑚
Example 2: The vector position of a particle varies in time according to the expression𝑟 =
(3𝑖^ − 6𝑡 2^
𝑗)𝑚.
a) Find expressions for the velocity and acceleration as functions of time.
b) Determine the particle’s position and velocity at𝑡 = 1𝑠.

Solution:

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Example 3: A particle starts with an initial velocity2.5 𝑚⁄𝑠 along the positive x direction and it
accelerates uniformly at the rate0.50 𝑚⁄𝑠 2 .
a) Find the distance travelled by it in the first two seconds.
b) How much time does it take to reach the velocity7.5 𝑚⁄𝑠?
c) How much distance will it cover in reaching the velocity7.5 𝑚⁄𝑠?
Solution:
1 1
a) ∆𝑥 = 𝑣⃗𝑥𝑖 𝑡 + 2 𝑎⃗𝑥 𝑡 2 = (2.5 𝑚⁄𝑠)(2𝑠) + 2 (0.50 𝑚⁄𝑠 2 )(2𝑠)2 = 6𝑚
⃗⃗𝑓 −𝑣
𝑣 ⃗⃗𝑖 7.5𝑚⁄𝑠−2.5𝑚⁄𝑠
b) 𝑡 = 𝑎
= 0.50𝑚⁄𝑠2
= 10𝑠
2 2
2 2 7.5 𝑚⁄𝑠 − 2.5 𝑚⁄𝑠
7.5 𝑚⁄𝑠 = 2.5 𝑚⁄𝑠 + 2(0.50 𝑚⁄𝑠 2 ∆𝑥) → ∆𝑥 = = 50𝑚
2(0.50 𝑚⁄𝑠 2 )

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Projectile Motion

The object which is given an initial velocity and afterwards follows a path determined by the
gravitational force acting on it is called projectile and the motion is called projectile motion.
Consider a body projected from a point 'O' with velocity 'u'. The point 'o' is called point of
projection and 'u' is called velocity of projection.

Velocity of Projection (u): the velocity with which the body projected.
Angle of Projection (α): The angle between the direction of projection and the horizontal plane
passing through the point of projection is called angle of projection.
Trajectory (OAB): The path described by the projectile from the point of projection to the point
where the projectile reaches the horizontal plane passing through the point of projection is called
trajectory. The trajectory of the projectile is a parabola.
Basic assumptions in projectile motion:
✓ The free fall acceleration (g) is constant over the range of motion and it is directed
downward.
✓ The effect of air resistance is negligible.
With the above two basic assumption the path of the projectile will be a down ward parabola.

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Example 1: A stone is thrown from the top of a building upward at an angle of 30° to the horizontal
with an initial speed of 20 𝑚⁄𝑠 if the height of the building is45𝑚.
a) How long does it take the stone to reach the ground?
b) What is the speed of the stone just before it strikes the ground?

Solution:

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Example 2: A projectile is fired in such a way that its horizontal range is equal to three times its
maximum height. What is the angle of projection?
Solution
𝑢2 𝑠𝑖𝑛2 𝜃 𝑢2 𝑠𝑖𝑛(2𝜃)
𝑅 = 3ℎ, ℎ = ,𝑅 =
2𝑔 𝑔
𝑢2 𝑠𝑖𝑛(2𝜃) 𝑢2 𝑠𝑖𝑛2 𝜃 4 4
→ = 3( ) → 𝑡𝑎𝑛𝜃 = 𝜃 = 𝑡𝑎𝑛−1 ( ) = 53°
𝑔 2𝑔 3 3
Example 3: A ball is thrown with an initial velocity of 𝑢⃗⃗ = (8𝑖 + 6𝑗) m/s. When it reaches thetop
of its trajectory, neglecting air resistance, what is its:
a) Velocity? b) Acceleration?
Solution
a) 𝑢⃗⃗ = 𝑢⃗⃗𝑥 = (8𝑖) 𝑚⁄𝑠
𝑎⃗ = −𝑔

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Uniform Circular Motion

Uniform Circular Motion is motion of objects in a circular path with a constant speed. Objects
moving in a circular path with a constant speed can have acceleration.

There are two ways in which the acceleration can occur due to:
✓ change in magnitude of the velocity
✓ change in direction of the velocity
For objects moving in a circular path with a constant speed, acceleration arises because of the
change in direction of the velocity.
Hence, in case of uniform circular motion:
✓ Velocity is always tangent to the circular path and perpendicular to the radius of the circular
path.
✓ Acceleration is always perpendicular to the circular path, and points towards the center of
the circle. Such acceleration is called the centripetal acceleration.

Period (T):- Time required for one complete revolution for a particle moving in a circle of radius
r with a constant speed

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Example 1: An athlete rotates a discus along a circular path of radius1.06𝑚. If the maximum speed
of the disc us is20 𝑚⁄𝑠, determine the magnitude of the maximum centripetal acceleration.
Solution:

Example 2: A tire 0.5𝑚 in radius rotates at a constant rate200 𝑟𝑒𝑣⁄𝑚𝑖𝑛. Find the speed and
acceleration of a small stone lodged in the tread of the tire.
Solution:

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