Speed Velocity and Acceleration

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April 2, 2025

Speed Velocity and Acceleration

Speed: The distance travelled by a particle per unit time is called speed. Speed is a scalar quantity. Speed = .

If s distance travelled by a particle in time t then speed of the particle is .

SI unit of speed is ms^-1 and the dimension is [LT ^-1].

Uniform speed: A particle or a body is said to be moving with uniform speed, if it covers equal distance in equal interval of time.

Variable speed: A particle or a body is said to be moving with variable speed, if it covers unequal distance in equal interval of time.

Average speed: The average speed of a body is defined as the ratio of the total distance travelled by it to the total time taken.

If S₁, S₂, S₃ are the distance travelled by a particle for time t₁, t₂, t₃ respectively, then the average speed of it is .

Instantaneous speed: The instantaneous speed at a point is deﬁned as the limit of the average speed over a path length that tends to become zero but always includes the point.

Let us consider a particle moves along a path ABCD and we want to calculate the instantaneous speed of it at point C. As the path s is very small, the corresponding time interval t also small.

The instantaneous speed v is thus given by v = = .

Velocity: The displacement of a particle per unit time is called velocity. Velocity is a vector quantity. Velocity = .

If is the displacement of a particle in time t then velocity of the particle is .

SI unit of velocity is ms^-1 and the dimension is [LT^-1].

Uniform velocity: A particle or a body is said to be moving with uniform velocity, if it covers equal displacement in equal interval of time.

Variable velocity: A particle or a body is said to be moving with variable velocity, if it covers unequal displacement in equal interval of time.

Average velocity: The average velocity of a body is defined as the ratio of the total displacement travelled by it to the total time taken.

If are the displacement travelled by a particle for time t₁, t₂ respectively, then the average velocity of it is .

Instantaneous velocity: The instantaneous velocity at a point at a certain instant of time is the limit of the average velocity over a displacement that tends to zero but always includes the point.

The magnitude of the instantaneous velocity is the instantaneous speed and its direction is along the tangent to the path drawn at the corresponding point.

Let us consider a particle moves along a path ABCD and we want to calculate the instantaneous velocity of it at point C. As the displacement Δ from point B to Dis very small, the corresponding time interval Δt also small.

The instantaneous speed v is thus given by = = .

SI unit of acceleration ms^-2and the dimension is [LT^-2].

CONCEPT ON VELOCITY:

The length of the minute hand of a clock is r cm. Calculate the magnitude of change in velocity of the tip of the minute hand from 12 noon to 12.15 pm.

The speed of the minute hand of a clock is v = = = cm

When the minute hand of a clock is at 12noon the direction of velocity () is towards +ve x axis. When the minute hand of a clock is at 12.15pm the direction of velocity ( ) is towards -ve y axis. Using vector subtraction method the change in velocity is ( – )

= = cm. [Magnitude of or is || = ||= ].

CONCEPT: 2. The half of the distance travelled by a car with speed and the other half with speed . Calculate the average speed of the car.

Let us consider the total distance is 2x. Time for travelling the distance x with speed is .

Time for travelling the distance x with speed is .

The average speed of the car is .

3. A train travels with uniform acceleration. When the front face of the train passes the electric post its speed is u and when the end face passes that post its speed is v. Calculate the speed of the train when the middle part of it passes that post.

Let the length of the train is L and acceleration is a. Then, or, .

The speed of the train when the half part of it passes that post is w. So,

Or,

A particle is moving with constant speed v in a circular path of radius r. It creates angle at the centre when moves from point A to B along the circumference of the circle. Find the change in velocity of the particle: At point A velocity of the particle is v_A (magnitude = v) and direction is along the tangent AC. Similarly at point B velocity of the particle is v_B (magnitude = v) and direction is along the tangent BD.

Draw a perpendicular AE on OB. Then EAF = as AE and FB are parallel and AC is the intersection then, CFB = . Therefore, the angle between v_A and v_B is .

So, change in velocity

of the particle is

Δv = [using vector subtraction rule( – )]

= v

= vsin

= 2v sin.

Average acceleration of the particle: The time taken to travel the distance AB is t, then = t = or, t =

Average acceleration = = = sin .

In the figure, the blocks A and B are of equal mass connected by a massless inextensible string over a pulley P which is frictionless. At any instant Block A moves on a horizontal surface with speed V_A and string creates angle with it. At that time B is moving with speed V_B. Find the relation between V_A and V_B.

Let the lengths of the sections of the string be AP = l₁ and PB = l₂ then the total length of the string is (l₁ + l₂).

In a further time t, let A moves to the right (point A^/) by x and B move down (point B^/) by y.

A^/C is perpendicular on AP.

Now A^/P PC = PA – AC = PA – AA^/ = l₁ – x and PB^/ = l₂ + y.

So, the total length of the string is A^/P + PB^/ = l₁ – x + l₂ + y = (l₁ + l₂ – x + y).

Therefore, l₁ + l₂ = l₁ + l₂ – x + y

Or, x = y

Differentiating both side of equation with respect to time t we get,

Or, =

V_A= V_B [The velocities of A and B are respectively V_A = and V_B = .]

A particle slides without friction from the highest point A of a vertical circle along a chord. If the particle starts from rest, show that the time of descent is independent of the chord chosen.

Let the radius of the circle is R. If the particle slides along the chord AB as shown in the figure, then the component of acceleration due to gravity along AB is gcos .

The length AB = 2AC = 2AOcos . = 2Rcos . [where is the angle made by AB with the vertical diameter.]

Using the equation s = a we get, 2Rcos . = gcos . or, t = .

So, the time of descent of the particle is independent of the chord chosen.

Acceleration: The rate of increase of velocity of an object with time is called acceleration.

Acceleration = .

SI unit of acceleration ms^-2and the dimension is [LT ^-2].

Retardation: The rate of decrease of velocity of an object with time is called retardation.

Motion (Jee advanced)

Three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side d at time

t = 0. Each of the particles moves with constant speed u. A always has its velocity along AB, B along BC and C along CA. Find the time when the particles meet each other.

Velocity of A is always along AB. Velocity of B is always along BC. So component of B along BA is vcos60° = . Thus, the separation AB decreases. The relative speed of A with respect to B along AB is v + = .

Time taken to reduce the distance AB (= d) is t = = .

Four boys A, B, C and D are initially at rest at four corners of a square of each side a. Each boys are moving with a uniform speed v in such a way that A always moves towards B, B always moves towards C, C always moves towards D and D always moves towards A. Where and at what time will they meet?

At any moment four boys are at the corner of a square whose sides are gradually decreases and finally meet at centre O. The velocity of A is v along AB and velocity of B is v along BC. So the velocity of B along A is vcos = 0 (as AB and BC are perpendicular)Thus, the separation AB decreases at a rate

of v + 0 = v. So, the time taken in reducing the separation AB from a to zero is t = .

In the figure, the pulley P moves to the right horizontally with a constant speed u. The downward speed of block B is V_B, and block A is moving on horizontal plane to the right with speed V_A. Find the relation between V_A and V_B.

At any instance of time, let the length of the string AP = l₁ and the length PB = l₂. In a further time t, let A moves to the right (point A^/) by x and B move down (point B^/) by y, while P moves to the right (point P^/) by ut. The velocities of A and B are respectively V_A = and V_B = .

As the length of the string must remain constant, AP + PB = A^/P^/ + P^/B^/

Or, l₁ + l₂ = (l₁ – x + ut) + (l₂ + y)

Or, x = ut + y

Differentiating both side of equation with respect to time t we get,

Or = +

V_A= u + V_B.

In the figure, the blocks A and B are of equal mass connected by a massless inextensible string over a pulley P which is frictionless. At any instance Block A moves on a horizontal surface with speed V_A and string creates angle with it. At that time B is moving with speed V_B. Find the relation between V_A and V_B.

Let the lengths of the sections of the string be AP = l₁ and PB = l₂ then the total length of the string is (l₁ + l₂).

In a further time t, let A moves to the right (point A^/) by x and B move down (point B^/) by y.

A^/C is perpendicular on AP.

Now A^/P ≈ PC = PA – AC = PA – AA^{/ cos} = l₁ – x cos and PB^/ = l₂ + y.

So, the total length of the string is A^/P + PB^/ = l₁ – x cos + l₂ + y = (l₁ + l₂ – x cos + y).

Therefore, l₁ + l₂ = l₁ + l₂ – x cos + y

Or, x cos = y

Differentiating both side of equation with respect to time t we get,

Or, =

V_{A cos}= V_B [The velocities of A and B are respectively V_A = and V_B = .]

The time taken by a particle to slide down an inclined plane is least along the line of greatest slope.

Let us consider an inclined plane of height h and angle of inclination is . A particle is released from the top (point A) of the inclined plane slide down along the greatest slope AB. The acceleration of the particle along AB is a = g sin and the travel time is t₁. Then using the relation for line AB is s = ut + at² we get, = g sin

Or, t₁ = = cosec .

Another particle is slides down from point A to C. The distance AC = AB sec = h cosec sec . Acceleration along AC is a = (g sin)cos and the travel time is t₂. Then using the relation for line AC is s = ut + at² we get,

hcosec sec = g sincos

Or, t₂ =

∴ t₂ =cosec sec = t₁ sec .

As sec >1 [as 0<<]

so, t_{2 > t1}

RELATIVE VELOCITY THEORY AND CONCEPTS

THEORY: Relative velocity: Let us consider and are the velocity of two particles A and B creates angleθ. The velocity of A with respect to B is = –. The angle between and – is (180- θ) and is the resultant of and – .

The magnitudes of is V_AB² = V_A² +V_B²+ 2V_AV_{B cos(180-θ)}

Or, V_AB² = V_A² +V_B²+ 2V_AV_{B cosθ}.

If creates angle φ with then, φ =

Or, tanφ = .

CONCEPTS ON RELATIVE VELOCITY IN ONE DIMENSION:

Two towns A and B are connected by a regular bus service with a bus leaving in either direction in every T minutes. A man cycling with speed of u in the direction from A to B notices that a bus goes pass him every t₁ minutes in the direction of his motion, and every t₂ minutes (t_1>t₂) in the opposite direction of his motion. What is the period T of the bus service and with what speed (assume constant) do the busses ply on the road?

Let us consider speed of each bus is v and the distance between the nearest two busses plying on either side is vT.

For bus going from town A to B:

Speed of bus with respect to cyclist is (v-u). As the cyclist notices that a bus goes pass him every t₁ minutes in the direction of his motion, then the separation between the busses is (v-u)t₁. So, vT = (v-u)t₁ —- (i)

For bus going from town B to A:

Speed of bus with respect to cyclist is (v+u). As the cyclist notices that a bus goes pass him every t₂ minutes in the opposite direction of his motion, then the separation between the busses is (v+u)t₂. So, vT = (v+u)t₂—- (ii)

From equations (i) and (ii) (v-u)t₁ = (v+u)t₂

Or, v = .

Using equation (i) vT = (v-u)t₁

∴T = .

A polish van moving with speed u on a highway fires a bullet at a thief’s car which is moving away with speed v in the same direction. If the muzzle speed of the bullet is w, with what speed does the bullet hit the thief’s car?

Muzzle speed is the speed of bullet with respect to the van. Speed of van or velocity of thief’s car are measured with respect to ground.

So, velocity of bullet with respect to van = velocity of bullet with respect to ground – velocity of van with respect to ground

W = velocity of bullet with respect to ground – u

Or, velocity of bullet with respect to ground = w + u.

Now, velocity of bullet with respect to thief’s car = velocity of bullet with respect to ground – velocity of thief’s car with respect to ground

So, velocity of bullet with respect to thief’s car = w + u – v.

On a two lane road, car A is moving with speed u. Two cars B and C approach car A in opposite direction with speed 2u. At a certain instant, when the distance AB = AC = d, driver of car B, decides to overtake A before C does. What minimum acceleration of car B is required to avoid accident?

At the instant when driver of car B, decides to overtake A the velocities of car A, B and C are respectively V_A = u, V_B = 2u and V_C = -2u [ -ve due to opposite direction].

Velocity of car B with respect to car A is V_BA = V_B – V_A = 2u – u = u.

Velocity of car C with respect to car A is V_CA = V_C – V_A = -2u – u = -3u.

Time required by C to cross A is t = = .

To avoid accident, car B must overtake car A in this time. Therefore, d = V_BAt + at²

Or, d = u() + a()²

So, a = .

RIVER – BOAT CONCEPT: In river boat concept V_R = absolute velocity of river or velocity of river with respect to ground,

V_B = absolute velocity of boat or velocity of boat with respect to ground and

V_BR = velocity of boatman with respect to river or velocity of boat in still water.

Let us consider boatman starts from point A from one bank of river with velocity V_BR as shown in figure. = –

Or, = + .

We can resolute all the velocities along x and y axis.

Absolute velocity of boat along x axis is V_Bx = – V_BR sinθ+ V_R = V_R – V_BRsinθ.

Absolute velocity of boat along y axis is V_By = V_BR cosθ+ V_R= V_{BR cosθ}

If D is the width of the river then time to cross the river is t = = ——- (i)

Drift or displacement of boatman along x axis when he reaches the other bank is x = V_Bxt = (V_R – V_{BR sinθ})t

= —–(ii)

Conditions:(i) When the boat crosses the river in shortest interval of time: From equation (i), time should be minimum if

θ = 0⁰ and that time is t = . I.e. boatman should drive the boat perpendicular to river (along AB).

(ii) When boatman wants to reach opposite point (point B) from where he started: From equation (ii), drift x = 0

Or, = 0.

So, =

Or, sinθ =

∴θ = ()

Therefore, to reach point B, boatman should row the boat at an angle θ = () ) upstream from AB.

(iii) Shortest path travelled by boatman: Path travelled by boatman when he reaches the opposite shore is S = .

(a) When<: From equation (ii), drift x = 0 at θ = ().

So, minimum path travelled by boatman S_min = D.

(b) When >: From equation (ii), drift x is minimum when = 0

Or, [] = 0

Or, [ – tanθ ] = 0

Or, – θ = 0

Or, = secθ

Or, sin θ=

∴θ = ().

Therefore minimum value of x is x_min = = = . [Using equation (ii)]

Shortest path travelled by boatman is S_min = = = .

EXAMPLE: The width of a river is D. A swimmer crosses the river and moves to the opposite point on the shore for time t₁ when there is no current in the river. He takes time t₂ to cross the river when river speed is v. Find the speed of river.

Let us consider the speed of swimmer in still water is u. so, D = ut₁.

When speed of river is v, then actual speed of swimmer is

So, D = t₂.

Then, D² = ()

Or, D² = ( – )

∴ v = D

RAIN – MAN CONCEPT:

(i) Man is moving on horizontal surface with constant speed:A man is moving on horizontal surface with constant speed V_M = u and rain is falling vertically with speed V_R = v. Find the velocity of rain with respect to man and at what angle with the vertical the man holds umbrella to protect him?

We know that = –

So, = = .

Rain is falling at an angle θ = () =() . Therefore the velocity of rain with respect to man is and the man holds umbrella at an angle θ = () with the vertical.

CONCEPTS ON SHORTEST DISTANCE: 1. Two ships A and B are at a distance d apart along the line in north-south direction. Ship A is situated to the north of ship B and is moving due west with a speed v. Ship Bis moving towards north with speed v. Find out the distance of closest approach between them and also calculate the time required to come to this position.

[As two velocity vectors have different origin, then we have to represent those vectors from a common origin.]

Velocity of ship B with respect to ship A is = – or, = = 2v (along BC) and creates angle φ with . So, tanφ = = = or,φ = 30⁰

Let AC is perpendicular on BC and AC is the shortest distance between two ships. So, AC = AB sinφ = d sin = .

The time required to come to this position is t = = = = = . [As we know the velocity along BC, we consider the time to move from B to C.]

Two cars A and B are moving with speed u and v respectively along two mutually perpendicular straight lines as shown in figure. What is the minimum distance between two cars?

Let us consider two cars are at a distance a and b respectively from point O and after time t the distance between two cars is minimum when they are at position A^/ and B^/ respectively. If L is the minimum distance between two cars then, = +