Cheat Sheet: Kinematics | Physics Class 11 - NEET PDF Download

 Page 1


Motion in a Straight Line
Key Concepts
• Displacement (s): Change in position, s = x
f
-x
i
(m).
• Velocity (v): Rate of change of displacement, v = ds/dt (m/s).
• Acceleration (a): Rate of change of velocity, a = dv/dt (m/s
2
).
• Types of Motion:
– Uniform motion: Constant velocity.
– Uniformly accelerated motion: Constant acceleration.
Equations of Motion (Constant Acceleration)
• v = u+at
• s = ut+
1
2
at
2
• v
2
= u
2
+2as
• s =
(
u+v
2
)
t
where u = initial velocity, v = ?nal velocity, a = acceleration, s = displacement, t = time.
Graphical Analysis
Graph Interpretation
Position vs. Time
(x-t)
Slope = velocity; curvature indicates acceleration.
Velocity vs. Time
(v-t)
Slope = acceleration; area under curve = displace-
ment.
Acceleration vs.
Time (a-t)
Area under curve = change in velocity.
Motion in a Plane
Key Concepts
• Position Vector: r = xi+yj (m).
• Velocity Vector: v =
dr
dt
= v
x
i+v
y
j (m/s).
• Acceleration Vector: a =
dv
dt
= a
x
i+a
y
j (m/s
2
).
• Components: Motion in x and y directions is independent (e.g., in projectile motion, a
x
= 0,
a
y
=-g).
1
Kinematics	Cheat	Sheet
(EduRev)
Page 2


Motion in a Straight Line
Key Concepts
• Displacement (s): Change in position, s = x
f
-x
i
(m).
• Velocity (v): Rate of change of displacement, v = ds/dt (m/s).
• Acceleration (a): Rate of change of velocity, a = dv/dt (m/s
2
).
• Types of Motion:
– Uniform motion: Constant velocity.
– Uniformly accelerated motion: Constant acceleration.
Equations of Motion (Constant Acceleration)
• v = u+at
• s = ut+
1
2
at
2
• v
2
= u
2
+2as
• s =
(
u+v
2
)
t
where u = initial velocity, v = ?nal velocity, a = acceleration, s = displacement, t = time.
Graphical Analysis
Graph Interpretation
Position vs. Time
(x-t)
Slope = velocity; curvature indicates acceleration.
Velocity vs. Time
(v-t)
Slope = acceleration; area under curve = displace-
ment.
Acceleration vs.
Time (a-t)
Area under curve = change in velocity.
Motion in a Plane
Key Concepts
• Position Vector: r = xi+yj (m).
• Velocity Vector: v =
dr
dt
= v
x
i+v
y
j (m/s).
• Acceleration Vector: a =
dv
dt
= a
x
i+a
y
j (m/s
2
).
• Components: Motion in x and y directions is independent (e.g., in projectile motion, a
x
= 0,
a
y
=-g).
1
Kinematics	Cheat	Sheet
(EduRev)
Projectile Motion
• Initial Velocity: u at angle ? to horizontal; components: u
x
= ucos?, u
y
= usin?.
• Equations (assuming g = 9.8 m/s
2
downward):
– Horizontal: x = ucos?·t, v
x
= ucos?.
– Vertical: y = usin?·t-
1
2
gt
2
, v
y
= usin?-gt.
• Key Quantities:
– Time of ?ight: T =
2usin?
g
.
– Maximum height: H =
u
2
sin
2
?
2g
.
– Range: R =
u
2
sin2?
g
.
Relative Motion
• Relative Velocity: v
AB
= v
A
-v
B
.
• Application: Motion of one object as observed from another (e.g., river-boat problems).
Key Formulas and Concepts
• Average Velocity: v
avg
=
? r
? t
.
• Average Acceleration: a
avg
=
? v
? t
.
• Uniform Circular Motion:
– Centripetal acceleration: a
c
=
v
2
r
(m/s
2
).
– Angular velocity: ? =
v
r
(rad/s).
• Projectile Motion Symmetry: Time to reach max height =
T
2
, range maximized at ? = 45
?
.
Solved Examples
1. Motion in a Straight Line: A car accelerates uniformly from rest at 2 m/s
2
for 5 s, then moves
with constant velocity for 10 s. Find the total distance traveled.
• Solution:
– Phase 1 (accelerated motion, 0 to 5 s):
– u = 0, a = 2 m/s
2
, t = 5 s.
– v = u+at = 0+2·5 = 10 m/s.
– s
1
= ut+
1
2
at
2
= 0+
1
2
·2·5
2
= 25 m.
– Phase 2 (constant velocity, 5 to 15 s):
– v = 10 m/s, t = 10 s, s
2
= v·t = 10·10 = 100 m.
– Total distance = s
1
+s
2
= 25+100 = 125 m.
– Answer: 125 m.
2. Projectile Motion: A projectile is launched with a velocity of 20 m/s at 60
?
to the horizontal.
Find the time of ?ight, maximum height, and range (g = 10 m/s
2
).
• Solution:
2
Page 3


Motion in a Straight Line
Key Concepts
• Displacement (s): Change in position, s = x
f
-x
i
(m).
• Velocity (v): Rate of change of displacement, v = ds/dt (m/s).
• Acceleration (a): Rate of change of velocity, a = dv/dt (m/s
2
).
• Types of Motion:
– Uniform motion: Constant velocity.
– Uniformly accelerated motion: Constant acceleration.
Equations of Motion (Constant Acceleration)
• v = u+at
• s = ut+
1
2
at
2
• v
2
= u
2
+2as
• s =
(
u+v
2
)
t
where u = initial velocity, v = ?nal velocity, a = acceleration, s = displacement, t = time.
Graphical Analysis
Graph Interpretation
Position vs. Time
(x-t)
Slope = velocity; curvature indicates acceleration.
Velocity vs. Time
(v-t)
Slope = acceleration; area under curve = displace-
ment.
Acceleration vs.
Time (a-t)
Area under curve = change in velocity.
Motion in a Plane
Key Concepts
• Position Vector: r = xi+yj (m).
• Velocity Vector: v =
dr
dt
= v
x
i+v
y
j (m/s).
• Acceleration Vector: a =
dv
dt
= a
x
i+a
y
j (m/s
2
).
• Components: Motion in x and y directions is independent (e.g., in projectile motion, a
x
= 0,
a
y
=-g).
1
Kinematics	Cheat	Sheet
(EduRev)
Projectile Motion
• Initial Velocity: u at angle ? to horizontal; components: u
x
= ucos?, u
y
= usin?.
• Equations (assuming g = 9.8 m/s
2
downward):
– Horizontal: x = ucos?·t, v
x
= ucos?.
– Vertical: y = usin?·t-
1
2
gt
2
, v
y
= usin?-gt.
• Key Quantities:
– Time of ?ight: T =
2usin?
g
.
– Maximum height: H =
u
2
sin
2
?
2g
.
– Range: R =
u
2
sin2?
g
.
Relative Motion
• Relative Velocity: v
AB
= v
A
-v
B
.
• Application: Motion of one object as observed from another (e.g., river-boat problems).
Key Formulas and Concepts
• Average Velocity: v
avg
=
? r
? t
.
• Average Acceleration: a
avg
=
? v
? t
.
• Uniform Circular Motion:
– Centripetal acceleration: a
c
=
v
2
r
(m/s
2
).
– Angular velocity: ? =
v
r
(rad/s).
• Projectile Motion Symmetry: Time to reach max height =
T
2
, range maximized at ? = 45
?
.
Solved Examples
1. Motion in a Straight Line: A car accelerates uniformly from rest at 2 m/s
2
for 5 s, then moves
with constant velocity for 10 s. Find the total distance traveled.
• Solution:
– Phase 1 (accelerated motion, 0 to 5 s):
– u = 0, a = 2 m/s
2
, t = 5 s.
– v = u+at = 0+2·5 = 10 m/s.
– s
1
= ut+
1
2
at
2
= 0+
1
2
·2·5
2
= 25 m.
– Phase 2 (constant velocity, 5 to 15 s):
– v = 10 m/s, t = 10 s, s
2
= v·t = 10·10 = 100 m.
– Total distance = s
1
+s
2
= 25+100 = 125 m.
– Answer: 125 m.
2. Projectile Motion: A projectile is launched with a velocity of 20 m/s at 60
?
to the horizontal.
Find the time of ?ight, maximum height, and range (g = 10 m/s
2
).
• Solution:
2
– u = 20 m/s, ? = 60
?
, u
x
= 20cos60
?
= 10 m/s, u
y
= 20sin60
?
= 10
v
3 m/s.
– Time of ?ight: T =
2usin?
g
=
2·20·sin60
?
10
=
2·20·
v
3/2
10
= 2
v
3˜ 3.46 s.
– Maximum height: H =
u
2
sin
2
?
2g
=
20
2
·(
v
3/2)
2
2·10
=
400·3/4
20
= 15 m.
– Range: R =
u
2
sin2?
g
=
20
2
·sin120
?
10
=
400·
v
3/2
10
= 20
v
3˜ 34.64 m.
– Answer: Time of ?ight ˜ 3.46 s, height = 15 m, range ˜ 34.64 m.
3. Relative Motion: A boat moves at 5 m/s in still water, crossing a river 100 m wide ?owing at 3
m/s. If the boat is directed perpendicular to the current, ?nd the time to cross and the downstream
drift.
• Solution:
– Boatvelocity: v
b
= 5m/s(y-direction,acrossriver); rivervelocity: v
r
= 3m/s(x-direction,
downstream).
– Time to cross: t =
width
v
b
=
100
5
= 20 s (y-motion independent of river ?ow).
– Downstream drift: x = v
r
·t = 3·20 = 60 m.
– Answer: Time = 20 s, drift = 60 m.
4. Graphical Analysis: The velocity-time graph of an object shows a straight line from (0, 2) to (5,
7) m/s. Find the acceleration and distance traveled in 5 s.
• Solution:
– Acceleration: a = slope =
? v
? t
=
7-2
5-0
= 1 m/s
2
.
– Distance: Area under v-t graph = area of trapezoid =
1
2
·(2+7)·5 =
1
2
·9·5 = 22.5 m.
– Answer: Acceleration = 1 m/s
2
, distance = 22.5 m.
5. Uniform Circular Motion: A particle moves in a circle of radius 2 m with a speed of 4 m/s. Find
the centripetal acceleration and angular velocity.
• Solution:
– Centripetal acceleration: a
c
=
v
2
r
=
4
2
2
=
16
2
= 8 m/s
2
.
– Angular velocity: ? =
v
r
=
4
2
= 2 rad/s.
– Answer: a
c
= 8 m/s
2
, ? = 2 rad/s.
3