Kinematics
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Statics, Kinematics and Dynamics
Displacement and Distance
The shortest distance between two points is called displacement. It is a vector quantity directed along the direction of motion.
The length of actual path travelled by a body between two points is called distance. It is a scalar quantity.
Q. Can a body have zero displacement but non-zero distance travelled?
Yes. If a body returns to the same position after motion, then displacement is zero but distance travelled is non-zero.
Speed and Velocity
It is a vector quantity and its SI unit is m/s.
It is a scalar quantity and its SI unit is m/s.
Average Speed and Instantaneous Speed
Average speed of a body is defined as the ratio of total distance travelled to the total time taken.
The speed of a body at a particular instant of time on its path is called instantaneous speed.
Average Velocity and Instantaneous Velocity
Average velocity of a body is defined as the ratio of total displacement to the total time taken.
The velocity of a body at a particular instant of time is called instantaneous velocity.
Acceleration or Retardation
The rate of change of velocity is called acceleration.
It is a vector quantity and its SI unit is m/s2.
The rate of decrease of velocity is called retardation or deceleration or negative acceleration.
Equations of Motion with Uniform Acceleration
- v = u + at
- s = ut + ½at2
- v2 = u2 + 2as
- Snth = u + a2n − 12
Equations of Motion under Gravity
- v = u ± gt
- h = ut ± ½gt2
- v2 = u2 ± 2gh
- Snth = u ± g2n − 12
Distance Travelled in nth Second
We have,
Sn = un + ½an2
Sn−1 = u(n−1) + ½a(n−1)2
Now, Snth = Sn − Sn−1
= un + ½an2 − u(n−1) − ½a(n−1)2
= un + ½an2 − un + u − ½a(n−1)2
= u + ½a[n2 − (n−1)2]
= u + ½a[n2 − n2 + 2n − 1]
∴ Snth = u + a2n − 12
Equation of Motion — Graphical Treatment
Let us consider a body moving with constant acceleration a along a straight line with initial velocity u at t = 0. After time t, its final velocity becomes v.
From figure: OA = ED = u, EB = OC = v and AD = OE = t.
1. Derivation of v = u + at
Acceleration of body = slope of line AB
a = BDAD = EB − EDAD
a = v − ut
v − u = at
∴ v = u + at
2. Derivation of s = ut + ½at2
From the graph, a = BD/AD = BD/t
∴ BD = at
Displacement (s) = area of trapezium OABE
= area of △ADB + area of rectangle OADE
= ½(BD × AD) + (AD × ED)
= ½(at × t) + tu
∴ s = ut + ½at2
3. Derivation of v2 = u2 + 2as
We have, a = BD/AD = (EB − ED)/AD = (EB − ED)/AB
or, AD = EB − EDa
Now, displacement (s) = area of trapezium OABE
s = ½(EB + OA) × AD
= ½(v + u) × v − ua
= v2 − u22a
v2 − u2 = 2as
∴ v2 = u2 + 2as
Numericals — Linear Motion and Motion under Gravity
Q. 1(C): Two objects meet near a tower
An object is dropped from the top of a tower of height 156.8 m and, at the same time, another object is thrown vertically upward with velocity 78.1 m/s from the foot of the tower. When and where do the objects meet?
Let A be the top of tower and B be its foot. Let C be the point where both objects meet.
For dropped object:
x = ut + ½gt2
x = ½gt2 …(i)
For object thrown vertically upward:
h − x = 78.1t − ½gt2 …(ii)
Using (i) in (ii):
h − ½gt2 = 78.1t − ½gt2
h = 78.1t
t = 156.8 / 78.1 = 2 sec
Putting t = 2 sec in (i): x = ½ × 10 × (2)2 = 20 m
Hence, they meet 20 m below the top of the tower after 2 sec.
Q. 1(B): Two balls dropped from a 200 m tower
A ball is dropped from the top of a tower 200 m high. After 1 sec another ball is dropped with 20 m/s from the top of the tower. When and where do they meet?
Let A be the top of the tower, B the foot and C the point where both objects meet.
Case I: x = ½gt2 …(i)
Case II: x = 20(t − 1) + ½g(t − 1)2 …(ii)
From (i) and (ii):
½gt2 = 20(t − 1) + ½g(t − 1)2
5t2 = 20t − 20 + 5(t2 − 2t + 1)
5t2 = 20t − 20 + 5t2 − 10t + 5
15 = 10t
t = 1.5 sec
Putting t = 1.5 sec in (i):
x = ½ × 10 × (1.5)2 = 11.25 m
Hence, they meet 11.25 m below the top of tower after 1.5 seconds.
Q. 1(C): Car slowing with uniform retardation
A car travelling with a speed of 15 m/s is braked and it slows down with uniform retardation. It covers a distance of 88 m and its velocity reduces to 7 m/s. If the car continues to slow down at the same rate, after what further distance will it be brought to rest?
Case I: u = 15 m/s, s = 88 m, v = 7 m/s
v2 = u2 + 2as
72 = 152 + 2 × a × 88
49 − 225 = 176a
a = −1 m/s2
∴ Retardation = 1 m/s2
Case II: u = 7 m/s, v = 0, a = −1 m/s2, s = ?
02 = 72 + 2(−1)s
2s = 49
s = 24.5 m
Hence, the car will be brought to rest after a further distance of 24.5 m.
Q. 1(D): Distance covered in the last second of free fall
A ball falls freely from the top of a tower and during the last second of its fall it falls through 25 m. Find the height of the tower. (Ans: 45 m)
In last second it falls through Stth = 25 m.
Initial velocity, u = 0 m/s.
Stth = u + g2(2t − 1)
25 = 0 + (10/2)(2t − 1)
25/5 = 2t − 1
6 = 2t
t = 3 sec
Again, h = ut + ½gt2
= 0 × 3 + ½ × 10 × (3)2
= 45 m
Hence, height of the tower is 45 m.
Short Question: If displacement is proportional to square of time
If the displacement of a body is proportional to the square of time, state the nature of motion of the body.
By question, y ∝ t2
or, y = kt2
Velocity: dy/dt = d(kt2)/dt = 2kt
Acceleration: dv/dt = d(2kt)/dt = 2k
Hence, the body moves with constant acceleration.
Projectile Motion
Examples:
- Stone thrown horizontally.
- A bomb dropped from an aeroplane.
Projectile Fired at an Angle with Horizontal
The initial velocity u is resolved into two components:
- ux = u cosθ along horizontal.
- uy = u sinθ along vertical.
Let a projectile be thrown towards the sky from the ground with initial velocity u, making angle θ with the ground. Its velocity can be resolved into u cosθ along horizontal and u sinθ along vertical. Since acceleration due to gravity acts in the vertical direction only, horizontal velocity remains constant but vertical velocity is variable.
Let P(x, y) be any point which the projectile reaches after time t.
Motion along Horizontal
Using s = ut + ½at2:
x = uxt
x = u cosθ · t
t = xu cosθ …(i)
Motion along Vertical
Using s = ut + ½at2:
y = uyt + ½(−g)t2
y = u sinθ · t − ½gt2 …(ii)
Using equation (i) in equation (ii):
y = u sinθ × xu cosθ − ½g(x / u cosθ)2
y = x tanθ − g2u2cos2θx2 …(iii)
This is of the form y = ax + bx2, the equation of a parabola. Hence, the path of a projectile is parabolic.
Time of Flight, Maximum Height, Horizontal Range and Velocity
Time of Flight (T)
The time for which projectile remains in space is called time of flight.
Since the projectile returns to ground after time T:
h = uyT − ½gT2
0 = u sinθ · T − ½gT2
½gT2 = u sinθ · T
∴ T = 2u sinθg
Maximum Height (Hmax)
It is the greatest height to which a projectile rises above the point of projection.
At maximum height, vertical velocity becomes zero, i.e. vy = 0.
vy2 = uy2 − 2gHmax
0 = (u sinθ)2 − 2gHmax
∴ Hmax = u2sin2θ2g
Horizontal Range (R)
The horizontal distance covered by the projectile during its time of flight is called horizontal range.
R = horizontal velocity × time of flight
R = ux · T
R = u cosθ × 2u sinθg
∴ R = u2sin2θg
Maximum Horizontal Range (Rmax)
If sin2θ = 1, then sin2θ = sin90°, so θ = 45°.
Thus, the horizontal range is maximum if the projectile is fired at an angle of 45° with the horizontal.
Two Angles of Projection for the Same Horizontal Range
For angle θ:
R = u2sin2θ / g
Let R′ be the horizontal range when angle of projection is 90° − θ:
R′ = u2sin2(90° − θ) / g
= u2sin(180° − 2θ) / g
= u2sin2θ / g
∴ R′ = R
Thus, the two projection angles θ and 90° − θ give the same horizontal range.
Velocity at Any Instant
Numericals — Projectile Motion
Short Q. 2(A): Same horizontal range
A projectile fired at an angle 18° has certain horizontal range. State another angle of projection for the same horizontal range.
There are two angles of projection for the same horizontal range.
For θ = 18°:
R = u2sin(2 × 18°) / g = u2sin36° / g
Another angle = 90° − 18° = 72°
R′ = u2sin[2(90° − 18°)] / g = u2sin(180° − 36°) / g = u2sin36° / g
Hence, another angle of projection is 72° for the same horizontal range.
Short Q. 2(B): Effect of doubling initial velocity on Rmax
What is the effect on Rmax on doubling the initial velocity of a projectile?
Rmax = u2sin2(45°)/g = u2/g
If u′ = 2u:
R′max = (2u)2/g = 4u2/g = 4Rmax
So, on doubling the initial velocity, Rmax increases 4 times.
Short Q. 2(C): Angle when horizontal range equals maximum height
Find the angle of projection at which the horizontal range and Hmax of a projectile are equal.
Given, R = Hmax
u2sin2θ / g = u2sin2θ / 2g
sin2θ = sin2θ / 2
2sinθ cosθ = sin2θ / 2
4cosθ = sinθ
sinθ / cosθ = 4
tanθ = 4
θ = tan−1(4)
∴ θ = 75.96°
Long Q. 2(D): Baseball after 2 seconds
A batter hits a baseball so that it leaves the bat with an initial speed 37 m/s at an angle of 53°. Find the position of the ball and direction of its velocity after 2 seconds. Treat the baseball as projectile.
Initial velocity, u = 37 m/s
Angle of projection, θ = 53°; time, t = 2 sec
x = uxt = u cosθ · t = 37 × cos53° × 2 = 44.53 m
y = x tanθ − [g / (2u2cos2θ)]x2
= tan53°(44.53) − [10 / (2 × 372 × cos253°)](44.53)2
= 59.09 − 19.99
= 39.1 m
Thus, position of the ball is at P(44.53, 39.1).
For direction:
vx = u cosθ = 37 cos53° = 22.26
vy = u sinθ − gt = 37 sin53° − 10(2) = 9.549
θ = tan−1(vy/vx) = tan−1(9.549/22.26) = 23.21°
So, direction is θ = 23.21°.
Long Q. 2(E): Player runs to catch a baseball
A baseball is thrown towards a player with an initial velocity 20 m/s and 45° with the horizontal. At the moment the ball is thrown, the player is 50 m from the thrower. At what speed and direction must he run to catch the ball at the same height at which it was released?
Given, u = 20 m/s, θ = 45°
R = u2sin2θ / g = 202 × sin90° / 10 = 40 m
T = 2u sinθ / g = 2 × 20 × sin45° / 10 = 2√2 sec
Distance the player must run = 50 − 40 = 10 m
Speed = distance/time = 10 / (2√2) = 5/√2 = 3.53 m/s
Hence, the player must run towards the first player with speed 3.53 m/s.
Long Q. 2(F): Range, greatest height and least launch speed
A projectile is fired from ground level with a velocity of 500 m/s at 30° to the horizontal. Find the horizontal range and greatest vertical height to which it rises. What is the least speed with which it could be projected in order to achieve the same horizontal range? (g = 10 m/s2)
u = 500 m/s, θ = 30°, g = 10 m/s2
R = u2sin2θ / g = (500)2sin(2 × 30°) / 10
∴ R = 21651 m
Hmax = u2sin2θ / 2g = (500)2sin230° / (2 × 10)
∴ Hmax = 3125 m
For the least speed, R = uleast2sin2θ / g. Here u is least when sin2θ is maximum, and maximum sin2θ = 1.
R = uleast2 / g
uleast = √(Rg) = √(21651 × 10) = √216510
∴ uleast = 465.30 m/s.
Horizontal Projectile
Consider a body thrown horizontally from the top of a tower of height h with velocity u. Suppose after time t it reaches point P(x, y) at depth y and horizontal distance x from the point of projection. Since it is thrown horizontally, its initial vertical velocity uy = 0 and acceleration due to gravity acts in the vertical direction only; the horizontal velocity remains constant.
Motion along Horizontal
x = uxt
x = ut
t = x/u …(i)
Motion along Vertical
y = uyt + ½gt2
y = ½gt2 …(ii)
Using equation (i) in (ii):
y = ½g(x/u)2
y = g2u2x2 …(iii)
This is of the form y = bx2; hence the path of a horizontal projectile is also parabolic.
Time of Flight (T)
h = uyT + ½gT2
h = ½gT2
∴ T = √(2h/g)
Horizontal Range (R)
R = horizontal velocity × time of flight
R = uxT
∴ R = u√(2h/g)
Special Cases (I and II)
Case I
Time of flight:
h = −uyT + ½gT2
h = −u sinθ · T + ½gT2
Horizontal range: R = uxT = u cosθ · T
Case II
Time of flight:
h = uyT + ½gT2
h = u sinθ · T + ½gT2
Horizontal range: R = uxT = u cosθ · T
Velocity at Any Instant
Horizontal velocity at any instant: vx = ux = u
Vertical velocity at any instant: vy = uy + gt = gt
v = √(vx2 + vy2) = √(u2 + g2t2)
tanθ = vy/vx = gt/u
∴ θ = tan−1(gt/u)
Numericals — Horizontal and Elevated Projection
Long Q. 3(A): Aeroplane diving at 37°
An aeroplane diving at an angle of 37° with the horizontal drops a mail bag at a height of 730 m. The projectile hits the ground 5 sec after being released. What is the speed of the aircraft?
h = uyt + ½gt2
730 = u sin37° × 5 + ½ × 10 × 52
730 − 125 = 5u sin37°
121 = u sin37°
u = 121/sin37°
∴ u = 201.05 m/s. Hence, speed of the aircraft is 201.05 m/s.
Long Q. 3(B): Suitcase dropped from an aeroplane
An aeroplane is flying with a velocity of 90 m/s at an angle of 23° above the horizontal. When the plane is 114 m directly above a dog standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? Ignore air resistance.
h = −uyT + ½gT2
114 = −90 sin23° · T + ½ × 10 × T2
5T2 − 35.16T − 114 = 0 …(i)
Comparing with ax2 + bx + c = 0:
a = 5, b = −35.16, c = −114
T = [−b ± √(b2 − 4ac)] / 2a
= [35.16 ± √((−35.16)2 − 4(5)(−114))] / 10
= [35.16 ± √3516.63] / 10
= [35.16 ± 59.3] / 10
Taking positive value: T = (35.16 + 59.3)/10 = 9.446 sec
∴ T = 9.4 sec
R = uxT = u cosθ · T = 90 cos23° × 9.4
∴ R = 782.55 m. Hence, the suitcase lands 782.55 m from the dog.
Long Q. 3(C): Horizontal projection from a 100 m tower
A body is projected horizontally from the top of a tower 100 m high with a velocity of 9.8 m/s. Find the velocity with which it hits the ground.
Given, h = 100 m, ux = u = 9.8 m/s
T = √(2h/g) = √(2 × 100 / 10) = √20
∴ T = 4.47 sec
Velocity at any instant:
v = √(vx2 + vy2)
= √[u2 + (gt)2]
= √[(9.8)2 + (10 × 4.47)2]
= √2094.13
∴ velocity = 45.76 m/s
Moreover, tanθ = vy/vx = gt/u
θ = tan−1[(10 × 4.47)/9.8]
θ = tan−1(4.56)
∴ θ = 77.63°. Hence, speed at impact is 45.76 m/s and direction is 77.63°.
Relative Velocity
Consider two bodies A and B moving with velocities VA and VB inclined at an angle θ.
Magnitude of VAB
Reverse V⃗B and complete the parallelogram; the diagonal represents V⃗AB.
Using vector addition law:
VAB = √[VA2 + VB2 + 2VAVBcos(180° − θ)]
VAB = √[VA2 + VB2 − 2VAVBcosθ]
Case I: Two Bodies in Same Direction (θ = 0°)
VAB = √[VA2 + VB2 − 2VAVBcos0°]
= √[(VA − VB)2]
∴ VAB = VA − VB
Hence, when two bodies are in the same direction, relative velocity is VA − VB or decreases.
Case II: Two Bodies in Opposite Directions (θ = 180°)
VAB = √[VA2 + VB2 − 2VAVBcos180°]
= √[VA2 + VB2 + 2VAVB]
= √[(VA + VB)2]
∴ VAB = VA + VB
Hence, when two bodies are in opposite directions, relative velocity is VA + VB or increases.
Numericals — Relative Velocity
Long Q. 4(A): Swimmer crossing a 600 m river
A man wishes to swim across a river 600 m wide. He can swim at the rate of 4 km/h in still water and the river flows at 2 km/h. In what direction must he swim to reach a point exactly opposite to the starting point, and when will he reach it?
Width of river, d = 600 m
Velocity of man, Vm = 4 km/h
Velocity of river, VR = 2 km/h
From the velocity triangle:
sinθ = VR/Vm = 2/4 = 1/2
θ = sin−1(1/2) = 30°
So, the man can swim by making angle (90° + 30°) = 120° with the flow of river.
Cross-river velocity:
V = Vmcosθ = 4 cos30° = 4 × √3/2 = 2√3 km/h
= (2 × 1000√3)/(60 × 60) = 0.96 m/s
v = d/t
t = d/v = 600/0.96 = 625 sec
Hence, after 625 seconds he will reach the point exactly opposite.
Long Q. 4(B): Swimmer downstream and upstream
A swimmer’s speed along the river (downstream) is 20 km/h and he can swim upstream at 8 km/h. Calculate the velocity of the stream and the swimmer’s possible speed in still water.
Let velocity of river = VR and velocity of swimmer = VS.
Downstream: VR + VS = 20 km/h …(i)
Upstream: VS − VR = 8 km/h …(ii)
Adding (i) and (ii):
2VS = 28
VS = 14 km/h
Putting VS = 14 in (i):
VR + 14 = 20
VR = 6 km/h
Thus, velocity of stream is 6 km/h and possible speed of the swimmer is 14 km/h.
Additional handwritten solution on PDF page 32
The final page repeats the 600 m river-crossing situation with Vm = 4 km/h and VR = 2 km/h, using a velocity-triangle approach.
cosθ = VR/Vm = 2/4 = 1/2 = cos60°
∴ θ = 60°
Thus, he must run/swim (30° + 90°) = 120° with the water.
v = √(Vm2 − VR2) = √(42 − 22) = √12 km/h
= (√12 × 1000 m) / 60 min
v = 57.73 m/min
d = 600 m
t = d/v = 600/57.73 = 10.39 min
∴ t = 10.39 min.
Discussion
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