Motion in a plane – all derivations
Parallelogram law of vector addition
According to the parallelogram law of vector addition: If two vectors are considered to be the
adjacent sides of a parallelogram, then the resultant of the two vectors is given by the vector that is
diagonal passing through the point of contact of the two vectors.
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Consider two vectors A and B inclined at an angle θ as shown. Let their resultant be R .
Extend PQ and draw QR such that SR  QR .
In QRS
QR
 cos θ  QR  QS cosθ  B cosθ
QS
SR
 sinθ  SR  QS sinθ  B sinθ
QS
.........(i)
..........(ii)
In PSR
PS 
2
 PR    SR 
2
2
 R2  PQ  QR    SR 
2
2
 R2   A  B cosθ   B sinθ 
2
2
[using (i) and (ii)]
 R2  A 2  B2 sin2 θ  2ABcosθ  B2 sin2 θ


 R2  A 2  B2 sin2 θ  cos2 θ  2ABcosθ
 R2  A 2  B2  2ABcosθ
 R  A 2  B2  2AB cosθ


If R makes an angle α with A , then
SR
SR

PR PQ  QR
B sinθ
 tan α 
A  B cosθ
tan α 
 B sinθ 
 α  tan1 

 A  B cosθ 
Horizontal projection of projectile
Consider a projectile thrown with velocity u in horizontal direction from a height h as shown
Therefore,
ux  u, uy  0
a x  0, a y  g
Equation of path
x  ux t 
1 2
a x t  x  ut
2
y  uy t 
1 2
1
a y t  y   gt 2
2
2
t 
x
u
2
1 x
1g 2
y   g   
x
2 u
2u
Which is a quadratic equation. Thus, path of a projectile is parabolic in nature.
Time of flight
Total time for which the projectile remains in air is called time of flight.
1
 y  uy t  a y t 2
2
1
 h   0  t  gT 2
2
 T
2h
g
Horizontal range (R)
Maximum horizontal distance travelled by projectile.
1 2
axt
2
1
 R  uT   0  T 2
2
 x  ux t 
 Ru
2h
g
 h   0  t 
1
gT
2
Velocity at any instant
v x  ux  ax t
 vx  u
v y  uy  ay t
 v y  gt
As v  v 2x  v 2y
v  u2  g2 t 2
Angular projectile motion
Consider a body projected with velocity u at an angle θ with horizontal as shown.
Therefore,
u x  ucosθ, uy  u sinθ
ax  0
a y  g
Equation of path
x  ux t 
1 2
a x t  x  ucos θt
2
y  uy t 
1 2
1
a y t  y  u sinθt  gt 2
2
2
t 
x
ucos θ
 x  1  x 
 y  u sinθ 
  g

 ucos θ  2  ucosθ 
2
1
g
 2
 y  x tanθ   2
x
2
2  u cos θ 
Time of flight
Total time for which the projectile remains in air is called time of flight.
 y  uy t 
1 2
ay t
2
 0   u sinθ  T 
 T
2usinθ
g
1 2
gT
2
 y  0 when body
hits the ground
Maximum height attained
At maximum height v y  0
 0  uy  a y t
 0  usinθ  gt
usinθ
t
t
Putting this value in equation of y, we get
y  uy t 
1 2
ay t
2
 u sinθ  1  u sinθ 
 H  u sinθ 
  g

 g  2  g 
u2 sin2 θ 1 u2 sin2 θ
H

g
2
g
u2 sin2 θ
 H
2g
Horizontal range
1
 x  ux t  a x t 2
2
 2u sinθ  1
2
R  ucosθ 
  0 t
 g  2
2u2 sinθ cos θ
R 
g
 R
u2 sin2θ
g
Velocity at any instant
v x  ux  ax t  ucosθ
v y  uy  ay t  usinθ  gt
 v  v 2x  v 2y
2
v
 ucos θ 
2
  u sinθ  gt 
2
 v  u2 sin2 θ  u2 cos2 θ  g2 t 2  2u sinθgt


 v  u2 sin2 θ  cos2 θ  g2 t 2  2u sinθgt
 v  u2  g2 t 2  2usinθgt
Centripetal acceleration

Consider a body moving in a circle of radius r with velocity v. Let the position vector of body be r1
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when it is at P and r2 when it is Q. The velocity vector of body at P is v1 and Q it is v 2 . If angle
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
between r1 and r2 is θ then clearly angle between v1 and v 2 is also θ .
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Clearly | r1 |  | r2 |  r


Since the motion is uniform so, | v1 |  | v 2 |  v
Now in
QOP and CAB
OP OQ r


AB AC v
and QOP  CAB
 by SAS similarity rule
QOP CAB
so,
v v
v v  r 
 

r r
t r  t 
 ac 
v2
r
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