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Some Definition Distance: Distance between two points is the length of actual path travelled by the particle. It is a scalar quantity.
Unit : m(metre)
Displacement: Displacement is a vector drawn from the initial position (A) to the final position (B)
Unit : m(metre)
Velocity: Velocity is the rate of change of position vector.
Unit : ms–1 (metre per second)
Acceleration: Acceleration is the rate of change of velocity.
Unit : ms–2 (metre per second2)
Average speed:The average speed of an object can be defined as the total distance travelled by it in a particular interval of time. It can be calculated by dividing the total distance travelled by the total time taken.
average speed (avg) = Total distance covered/ Total time taken
Average velocity: Average velocity is a vector quantity. Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.The average velocity can be positive or negative depending upon the sign of the displacement. The SI unit of average velocity is meters per second (m/s or ms-1)
Acceleration is defined as the rate of change of velocity. It is denoted by ‘a’ and is measured in the units of m/s2. For a particular interval, the average acceleration is defined as the change in velocity for that particular interval. Unlike acceleration, the average acceleration is calculated for a given interval.
GRAPHICAL SOLUTION OF RECTILINEAR MOTION
v-t Curve: The area under the v-t curve measures the change in position x.
In the above graph,
Following details are obtained from the graph above:
The initial velocity of the body, u = OA
The final velocity of the body,v = BC
From the graph, we know that
BC = BD + DC
Therefore,v = BD + DC
v = BD + OA (since DC = OA)
v = BD + u (since OA = u)(Eqaution 1)
Now, since the slope of a velocity-time graph is equal to acceleration a, So,
a = slope of line AB
a = BD/AD
Since AD = AC = t, the above equation becomes:
BD = at (Equation 2)
Now, combining Equation 1 & 2 , the following is obtained
Derivation of second equation of motion:
Distance travelled (s) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD
s = (1/2 AB × BD) + (OA × OC)
Since BD = EA, the above equation becomes
s = (1/2 AB × EA) + (u × t)
As EA = at, the equation becomes
s = 1/2 × a × t + ut
Derivation of third equation of motion:
Characteristics of v-t graph: The velocity of an object is its speed in a particular direction. Two cars travelling at the same speed but in opposite directions have different velocities.
A velocity-time graph shows the speed and direction an object travels over a specific period of time. Velocity-time graphs are also called speed-time graphs.
When an object is moving with a constant velocity, the line on the graph is horizontal. When the horizontal line is at zero velocity, the object is at rest. When an object is undergoing constant acceleration, the line on the graph is straight but sloped.
Curved lines on velocity-time graphs also show changes in velocity, but not with a constant acceleration or deceleration. The diagram shows some typical lines on a velocity-time graph.
The steeper the line, the greater the acceleration of the object. The purple line is steeper than the green line because it represents an object with a greater acceleration.Notice that a line sloping downwards - with a negative gradient - represents an object with a constant deceleration (it is slowing down).Acceleration can be calculated by dividing the change in velocity (measured in metres per second) by the time taken for the change (in seconds). The units of acceleration are m/s/s or m/s2
Graphical representation of motion:
Addition of vectors:vector addition is commutative. So,
It means that the law of addition of vectors is independent of the order of vectors.
Subtraction of vectors:The process of subtracting one vector from another is equivalent to adding, vectorially, the negative of the vector to be subtracted. Suppose there are two vectors and , shown in figure (A) and we have to subtract and . It is just the same thing as adding vectors – to . The resultant is shown in figure (B).
Parallelogram law of vector addition:If two vectors acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal passing through the common tail of the two vectors.
CD is drawn perpendicular to the extended OA, from C. Let COD made by Vec R with Vec P be α.
From right angled triangle OCD,
OC2 = OD2 + CD2
= (OA + AD)2 + CD2
= OA2 + AD2 + 2.OA.AD + CD2 .....(1)
In Fig. Angle BOA = θ =Angle CAD
From right angled ∆ CAD,
AC2 = AD2 + CD2 ...(2)
Substituting (2) in (1)
OC2 = OA2 + AC2 + 2OA.AD ...(3)
CD = AC sin θ ...(4)
AD = AC cos θ ...(5)
Substituting (5) in (3) OC2 = OA2 + AC2 + 2 OA.AC cos θ
Substituting OC = R, OA = P,
OB = AC = Q in the above equation
R2 = P2 + Q2 + 2PQ cos θ(or)
R = root [ P2 + Q2 + 2PQ cos θ ] ...(6)
Equation (6) gives the magnitude of the resultant. From ∆ OCD,
tan α = CD / OD = CD/( OA +AD )
Substituting (4) and (5) in the above equation,
tan α = ( AC sin θ ) / (OA + AC cos θ )
(or) α = tan-1[Qsin θ / (P+Qcos θ)] ...(7)
Equation (7) gives the direction of the resultant
Relative Velocity: Relative velocity is used to denote the aircraft moving in the wind or boats moving through the water, etc. The velocity is measured within the object according to the observer. It can be measured using the use of an intermediate reference frame. In simpler terms, this can be the vector sum of the velocities. The relative velocity formula is expressed as
Where VABis the velocity with respect to A and B, VBC is the velocity with respect to B and C and V AC is the velocity with respect to A and C.Let us consider two objects, A and B moving with velocities Va and Vb with respect to a common stationary frame of reference, say the ground, a bridge or a fixed platform.
The velocity of the object A relative to the object B can be given as,
Vab = Va - Vb
Similarly, the velocity of the object B relative to that of object a is given by,
Vba = Vb - Va
From the above two expressions, we can see that
Vab = -Vba
Although the magnitude of both the relative velocities is equal to each other. Mathematically,
|Vab| = |Vba|
Resolution of vectors: The component of F in a direction making an angle θ is F cos θ.
The other component of F at right angles to F cos θ is F sin θ.
A vector directed at an angle with the co-ordinate axis, can be resolved into its components along the axes. This process of splitting a vector into its components is known as resolution of a vector.
Consider a vector R = Vector( OA) making an angle θ with X - axis. The vector R can be resolved into two components along X - axis and Y-axis respectively. Draw two perpendiculars from A to X and Y axes respectively. The intercepts on these axes are called the scalar components Rx and Ry.
Then, OP is Rx, which is the magnitude of x component of Vector R and OQ is Ry, which is the magnitude of y component of Vector R.
From ∆ OPA,
cos θ = OP/OA = Rx/R(or) Rx=Rcos θ
sin θ = OQ/OA = Ry/R (or) Ry=Rsin θ
R2 = Rx2 + Ry2
Also, Vector R can be expressed as Vector R = Rxi + Ryj where i and j are unit vectors.
In terms of Rx and Ry , θ can be expressed as θ = tan−1 [Ry/Rx]
Unit vector: A vector having unit magnitude is called a unit vector. It is also defined as a vector divided by its own magnitude. A unit vector in the direction of a vector A ,is written as Â and is read as 'A cap' or 'A caret' or 'A hat'. Therefore,
Thus, a vector can be written as the product of its magnitude and unit vector along its direction.
The geometric meaning of dot product says that the dot product between two given vectors and bis denoted by:
a⋅b = a ∣∣ b ∣ cos
Here, |a| and |b| are called as the magnitudes of vector a and b and θ is the angle between the vectors a and b.
If the two vectors are orthogonal, that is, the angle between them is 90, then a.b = 0 since cos 90 = 0.
If the two vectors are parallel to each other, then a.b =|a||b| since cos 0 = 1.
Dot Product Algebra Definition
The dot product algebra says that the dot product of the given two products - a = (a1,a2,a3,) and b = (b1,b2,b3)is given by:
a.b = (a1b1 + a2b2 + a3b3)
Dot Product of Two Vectors Properties
Given below are the properties of vectors:
Vector product or Cross product
The vector product or cross product of two vectors A and B is denoted by , A x B, and its resultant vector is perpendicular to the vectors A and B.The cross product is mostly used to determine the vector, which is perpendicular to the plane surface spanned by two vectors, whereas the dot product is used to find the angle between two vectors or the length of the vector. The cross product of two vectors, say A x B,is equal to another vector at right angles to both, and it happens in the three-dimensions.
If θ is the angle between the given two vectors A and B, then the formula for the cross product of vectors is given by:
A x B = |A| |B| sin θ
Or, Vector A x Vector B = ||Vector A || ||Vector B|| sin θn^
Here, Vector A, Vector B are two Vectors.
||Vector A || ||Vector B|| are the magnitudes of given vectors.
θ is the angle between two vectors and n^is the unit vector perpendicular to the plane containing the given two vectors, in the direction given by the right-hand rule.
Cross product of two vectors Formula
Consider two vectors,
A = ai + bj + ck
B = xi + yj + zk
We know that the standard basis vectors i,j and k satisfy the below-given equalities.
i x j = k and j x i = -k
j x k = i and k x j = -i
k x i = j and i x k = -j
Also, the anti-commutativity of the cross product and the distinct absence of linear independence of these vectors signifies that:
i x i = j x j = k x k = 0
A × B = (ai + bj + ck) × (xi + yj + zk) = ax(i × i) + ay(i × j) + az(i × k) + bx(j × i) + by(j × j) + bz(j × k) + cx(k × i) + cy(k × j) + cz(k × k)
By applying the above mentioned equalities,
A × B = ax(0) + ay(k) + az(-j) + bx(-k) + by(0) + bz(i) + cx(j) + cy(-i) + cz(0) = (bz – cy)i + (cx – az)j + (ay – bx)k
Cross product matrix:
We can also derive the formula for the cross product of two vectors using the determinant of the matrix as given below.
A = ai + bj + ck B = xi + yj + zk
A × B = (bz – cy)i – (az – cx)j + (ay – bx)k = (bz – cy)i + (cx – az)j + (ay – bx)k
Cross product of perpendicular Vector:
Cross product of two vectors is equal to the product of their magnitude, which represents the area of a rectangle with sides X and Y. If two vectors are perpendicular to each other, then the cross product formula becomes:
θ= 90 °
We know that, sin 90° = 1
Cross Product of parallel vector:
The cross product of two vectors are zero vectors if both the vectors are parallel or opposite to each other. Conversely, if two vectors are parallel or opposite to each other, then their product is a zero vector. Two vectors have the same sense of direction.
θ= 90 degrees
As we know, sin 0° = 0 and sin 90° = 1
Projectile Motion : When a particle is thrown obliquely near the earth’s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is called projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
Total Time of Flight: Resultant displacement (s) = 0 in Vertical direction. Therefore, by using the Equation of motion:
gt2 = 2(uyt – sy) [Here,uy = u sin θ and sy = 0]
i.e. gt2 = 2t × u sin θ
Time of flight :
Horizontal Range (OA) = Horizontal component of velocity (ux) × Total Flight Time (t)
R = u cos θ × 2u×sinθg
Therefore, in a projectile motion the Horizontal Range is given by (R):
Maximum horizontal range
Projectile on inclined plane:
Let the particle strike the plane at A so that OA is the range of the projectile on inclined plane. This initial velocity can be resolved into two components:
Time of Flight:- Let t be the time taken by the particle to go from A to B. In this time the displacement of the projectile to the plane is zero.
Hence, 0 = u sin (α-β) t - ½g β t2
=> t = 2u sin(α-β)/gcosβ
During time of flight, the horizontal velocity u cos α remains constant.
Hence, horizontal distance,
OB = (ucosα) t = 2u2sin(α-β)cosα/gcosβ
Now, OA = OB/cosβ = 2u2sin(α-β)cosα/gcosβ
The greatest distance of the projectile from the inclined plane is u2sin2 (α-β)/2gcosβ .
Circular motion is the movement of an object in a circular path.