1. MOTION IN A PLANE – (B) PROJECTILE & CIRCULAR MOTION
Position and Displacement Vectors
Velocity (General & Component Form)
Acceleration (General & Component Form)
Motion in a Plane with Constant Acceleration (Equations of Motion)
Relative Velocity (By Vector Algebra) and its Magnitude & Direction
Projectile Motion – Equations of a Projectile
Equation of path of Projectile, Time of Flight, Maximum Height & Range
Maximum Range and Range for Complement Angles of Projection
Circular Motion – Relation between Linear and Angular Velocity
Acceleration in Uniform Circular Motion – Direction and Magnitude
Acceleration in terms of Angular Speed and Frequency
Directions of r, v,  and acp
Created by C. Mani, Education Officer, KVS RO Silchar
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2. Position and Displacement VectorsX Y O
The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame is given by P r j y i
= x r j
+ y
where x and y are components of r along x- and y- axes. i x O Y X
Direction of vav
Let the particle move through the curve from P at time t to P′ at time t′. Then P′ Δy r′ P Δr
Displacement vector is r r′ -
Δr = r
It is directed from P to P′. or Δr
= (x′ i j
+ y′ ) - (x
+ y ) Δx i
= Δx j
+ Δy
where Δx = x′ - x and Δy = y′ - y
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3. Velocity
The average velocity vav of a particle is the ratio of the displacement to the corresponding time interval.
vav = Δt Δr = i Δx j
+ Δy = i Δx Δt j + Δy or i
= vx av
vav j
+ vy av
The direction of vav is same that of Δr.
The instantaneous velocity is given by the limiting value of the average velocity as the time interval approaches zero.
v = Δr Δt
lim
Δt→0 dr dt or
The meaning of limiting value is explained in “Motion in a straight line”.
The direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.
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4. Velocity in component form
lim
Δt→0 X Y O = i Δx Δt j + Δy
lim
Δt→0 v j vy P θ i vx = i Δx Δt +
lim
Δt→0 j Δy = i dx dt j + dy dx dt
vx = dy
vy =
where
and i
= vx v j
+ vy or
The magnitude of velocity v is
v =
vx2 + vy2
The direction of velocity v is
tan θ = vy vx
θ = tan-1 vy vx or
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5. Acceleration
The average acceleration a of a particle is the ratio of the velocity to the corresponding time interval.
aav = Δt Δv = i
Δ (vx j
+ vy ) = i
Δvx Δt j +
Δvy i
= ax
aav j
+ ay or
The instantaneous acceleration is given by the limiting value of the average acceleration as the time interval approaches zero.
a = Δv Δt
lim
Δt→0 dv dt or
The meaning of limiting value is explained in “Motion in a straight line”.
In one dimension, the velocity and the acceleration of an object are always along the same straight line (either in the same direction or in the opposite direction). However, for motion in two or three dimensions, they may have any angle between 0° and 180° between them.
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6. Acceleration in component formΔv Δt
lim
Δt→0 = i
Δvx Δt j +
Δvy
lim
Δt→0 = i
Δvx Δt +
lim
Δt→0 j
Δvy = i
dvx dt j +
dvy
where
dvx dt
ax =
d2x
dt2 =
and
dvy
ay =
d2y i
= ax a j
+ ay or
The magnitude of acceleration a is
a =
ax2 + ay2
The direction of acceleration a is
tan θ = ay ax
θ = tan-1 ay ax or
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7. MOTION IN A PLANE WITH CONSTANT ACCELERATION
Let the velocity of the object be v0 at time t=0 and v at time t.
Then
a =
v – v0
t – 0 = t or
v = v0 + at
In terms of components,
vx = v0x + ax t
vy = v0y + ay t
Let the position vector of the object be r0 at time t=0 and r at time t.
Then the displacement is the product of average velocity and time interval. t
r – r0 2
v + v0 = t 2 = v0
v0 + at +
r – r0 = v0t + ½ at2
r = r0 + v0t + ½ at2
In terms of components,
x = x0 + v0xt + ½ axt2
y = y0 + v0yt + ½ ayt2
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8. Relative Velocity - (By Vector Algebra)
When the velocity of an object is measured with respect to an object which is at rest or in motion, the velocity measured is known as relative velocity.
Consider two objects A and B moving with velocities vA and vB respectively.
To find the relative velocity of the object, say A with respect to the object B, the velocity -vB is superimposed on the object B so as to bring it to rest.
To nullify this effect, velocity -vB is superimposed on the object A also.
The resultant of vA and -vB gives the relative velocity vAB of the object A with respect to the object B. R
vAB = vA - vB vB
-vB vA P O
Mathematically, vAB = vA + (-vB) or
vAB = vA - vB
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9. When two objects are moving along the same straight line:
Magnitude and Direction of the Relative Velocity in terms of the Magnitudes and Angle θ between them B R
vAB = vA - vB
vAB
180°- θ α
-vB θ vB O vA A
The magnitude of vector vAB is
vAB =
vA2 + vB vA vB cos (180° - θ) or
vAB =
vA2 + vB vA vB cos θ)
The direction of vector vAB is
B sin θ
tan α =
A - B cos θ
When two objects are moving along the same straight line:
vAB = vA - vB (if they move in the same direction)
vAB = vA + vB (if they move in the opposite direction)
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10. PROJECTILE MOTION Projectile Projectile Motion
An object that is in flight after it being thrown or projected is called a projectile.
Projectile Motion
The motion of a projectile which is in flight after it being thrown or projected is called projectile motion.
It can be understood as the result of two separate, simultaneously occurring components of motion (along x- and y- axes).
The component along the horizontal direction (x- axis) is without acceleration.
The component along the vertical direction (y- axis) is with constant acceleration under the influence of gravity.
In our study, the air resistance is negligible and the acceleration due to gravity is constant over the entire path of the projectile.
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11. Equations of a Projectile Motion
Suppose that the projectile is launched with velocity v0 that makes an angle θ0 with the x-axis. X Y O
Acceleration acting on the projectile is due to gravity which is directed vertically downward: j
a = -g j
a = -g or
ax = 0, ay= -g v0
v0 sin θ0
The components of initial velocity v0 are:
v0x = v0 cos θ0 θ0
v0y = v0 sin θ0
v0 cos θ0
If the initial position is taken as the origin O, then
x0 = 0, y0= 0
becomes
x = x0 + v0xt + ½ axt2
x = v0xt = (v0 cos θ0)t
and
y = y0 + v0yt + ½ ayt2
becomes
y = (v0 sin θ0)t - ½ gt2
The components of velocity at time t are:
vx = v0x + ax t
becomes
vx = v0x = v0 cos θ0
vy = v0y + ay t
and
becomes
vy = v0 sin θ0 - gt
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12. j
a = -g X Y O v0 θ0
v0 cos θ0
v0 sin θ0
v0 sin θ0 - gt
-(v0 sin θ0 – gt) v
-θ0
-v0 sin θ0 vt β
Note:
The horizontal component of velocity remains constant throughout the motion.
But, the vertical component reduces to zero at its peak of the path and again increases in the opposite direction.
The magnitude of velocity of the projectile at an instant ‘t’ is given by
vt =
v02 cos2 θ0 + (v0 sin θ0 – gt)2
The direction of velocity of the projectile at that instant ‘t’ is given by
v0 sin θ0 - gt
tan β =
v0 cos θ0
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13. Equation of path of a projectile
The shape of the path of a projectile can be found by mathematical equation.
x = v0xt = (v0 cos θ0)t
From
we get
v0 cos θ0 x
t =
y = (v0 sin θ0)t - ½ gt2
becomes
y = (v0 sin θ0)
- ½ g
v0 cos θ0 x 2
On simplification,
y = (tan θ0) - x
2 (v0 cos θ0)2 g x2
Since g, θ0 and v0 are constants, the above equation is in the form of
where a = tan θ0 and b =
2 (v0 cos θ0)2 g
y = ax - bx2
The above equation is the equation of a parabola.
Therefore, the path of the projectile is a parabola.
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14. Time to reach Maximum Height and Time of Flight of a Projectile
Let tm be the time taken for the projectile to reach its maximum height and Tf be the total time of flight of the projectile.
At the point of maximum height and at t = tm , vy = 0.
vy = v0 sin θ0 - gt
becomes
0 = v0 sin θ0 - gtm or
v0 sin θ0 g
tm =
At t = Tf , y = 0.
y = (v0 sin θ0)t - ½ gt2
becomes
0= (v0 sin θ0)Tf - ½ gTf2 or
2 v0 sin θ0 g
Tf =
Note that Tf = 2 tm because of the symmetric nature of the parabolic path.
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15. Maximum Height of a Projectile
Let hm be the maximum height of the projectile after time tm.
y = (v0 sin θ0)t - ½ gt2
becomes
hm = (v0 sin θ0)
- ½ g 2
v0 sin θ0 g or
v02 sin2 θ0 2g
hm =
Aliter:
At hm (the maximum height of the projectile), vy = 0.
vy2 = v02 sin2 θ0 – 2gy
becomes
02 = v02 sin2 θ0 – 2ghm or
v02 sin2 θ0 2g
hm =
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16. Range of a Projectile
Let R be the Range of the projectile after time Tf (Time of flight). It is the horizontal distance covered by the projectile from its initial position (0,0) to the position where it passes y = 0.
becomes
x = v0xt = (v0 cos θ0)t
R = (v0 cos θ0) Tf or
R = (v0 cos θ0)
2 v0 sin θ0 g or
v02 sin 2θ0 g
R =
Note that the range will be maximum for the maximum value of sin.
i.e. when sin 2θ0 = 1. This is possible when θ0 is 45°.
v02 g
Rm =
Therefore, the maximum horizontal range is
When θ0 is 45°,
v02 4g
hm,45° =
and
Rm = 4 hm,45°
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17. Range of a Projectile is same for complement angles of projectionX Y O
v02 sin 2θ0 g
R = v0 v0 v0 α α
45°
Rmax
For angles, (45° + α) and (45° - α), 2θ0 is (90° + 2α) and (90° - 2α) respectively.
The values of sin (90° + 2α) and sin (90° - 2α) are the same and are equal to cos 2α. Therefore, ranges are equal for elevations which exceed or fall short of 45° by equal amounts of α.
In other words, for complement angles of elevation, the ranges will be the same.
i.e. for θ0 and (90° - θ0) the values of sin 2θ0 and sin (180° - 2θ0) are the same.
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18. UNIFORM CIRCULAR MOTIONv’
When a body moves with constant speed on a circular path, it is said to have uniform circular motion.  P’
A particle P moves on a circle of radius vector r with uniform angular velocity . r' Δ v Δr Δs O Δ
Linear velocity v is constant in magnitude but changes its direction continuously.
 The particle experiences acceleration. r P
In case of non-uniform circular motion, the particle experiences acceleration due to change in both speed and direction.
When the particle moves from P to P’ in time Δt = t’ – t, the line OP (radius vector) moves through an angle Δθ. Δθ is called ‘angular displacement’.
The velocity vector v turns through the same angle Δθ and becomes v’.
The linear displacement PP’ is Δr. The linear distance Δs is the arc PP’.
The angular velocity is the rate of change of angular displacement.
 = Δθ Δt
lim
Δt→0
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19. Relation between Linear and Angular Velocity
The linear velocity is the rate of change of linear displacement.
|v| = Δs Δt
lim
Δt→0
But Δs = r Δθ
|v| =
r Δθ Δt
lim
Δt→0 
|v| = Δθ Δt
lim
Δt→0 r or
|v| = r
||
v =  x r
, r and v are mutually perpendicular to each other and  is perpendicular to the plane containing r and v.
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20. Acceleration in Uniform Circular Motion
Direction of acceleration of a particle in a uniform circular motion P’ v’  β β β β Δv Δ Δv Δ Δv Δ r' v P Δv Δv Δ Δ Δr V V Δs V’ V V v’ v O Δ Δ r
As Δt→0, Δ→0° and β→90°. It means the angle between Δv and v, i.e. β increases and approaches 90°. i.e. Δv becomes perpendicular to v.
r is perpendicular to v. And Δv is also perpendicular to v.
 Δv is acting along -r. (Note the negative sign)
Since acceleration is the rate of change of velocity, therefore it acts in the direction of Δv. Or it acts in the direction along the radius and towards the centre O. Hence, the acceleration is called ‘centripetal acceleration’.
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21. Magnitude of acceleration of a particle in a uniform circular motionv’  r' v P Δv A Δ B Δr Δs v’ v O Δ Δ r P
The two isosceles triangles OPP’ and PAB are similar triangles. = AB
PP’ PA OP
|v|
|Δv| =
|Δr|
|r|
|v|
|Δv| =
|Δr|
|r|  or or
|a|=
lim
Δt→0 Δt
|Δr|
|v|
|r|
|a|=
lim
Δt→0 Δt
|v|
|Δr|
|r| x
|Δv|
|a|= Δt
lim
Δt→0 or 
|a|=
|v|
|r|
|a|=
|v|2
|r|
acp = v2 r or or or
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22. Centripetal acceleration can be expressed in terms of angular speed.
acp = v2 r
But v = r 
acp =
(r)2 r or
acp = 2r
Centripetal acceleration in terms of frequency can be expressed as:
 = 2πν 
acp = 2r
becomes
acp = (2πν)2 r or
acp = 4π 2 ν2 r
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23. Directions of r, v,  and acp
The relative directions of various quantities are shown in the figure. O  v r
acp P
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24. Acknowledgement 1. Physics Part I for Class XI by NCERT ] Home End
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