Motion

A has greater velocity.

Explanation:

The slope of distance and time graph gives the velocity.

Since,

Velocity =

The formula for slope of any graph line is the ratio of change in value of y-axis to the change in value of x-axis.

x = t

y = s

so, slope =

Thus from the above diagram, if we estimate the change of distance on y-axis to the change of time on x-axis of both A and B, then we will find that the slope corresponding to A is greater as change in distance in A is more than that of B, if we keep the change in time constant.

Hence A has higher velocity than B

A man climbing a ladder

Explanation:

Distance is a scalar quantity (it has magnitude only) while displacement is a vector quantity (it has magnitude and direction both)

Hence if we take any example considering a motion in which direction does not change throughout the motion and we will always get

Distance = Displacement.

Two examples of vector quantities are:

• Velocity

• acceleration

Since in both the cases both the magnitude as well as the direction plays a role in assessing its true value.

Average speed of a body is defined as the total distance travelled divided by the total time taken to cover the distance.

Average speed =

Let us consider a car is travelling 2 Km in 30 minutes and then next 3 Km at 20 minutes. Then the average speed of the car is

Average speed = = 0.1 Km/min

SI unit of angular momentum is Radian per second (rad-s^-1)

Explanation:

Since,

Angular Velocity =

SI unit of

• is radians

• is seconds

Hence

SI unit of Angular velocity is radians per second (rad s^-1)

Two uses of Graphical representation of motion are

• Just by looking at the shape of the curve we can tell the nature of motion (i.e. uniform motion, non uniform, uniformly accelerated etc.)

For Example,

• if the distance – time graph is having straight line passing through origin then we can say that the motion is uniform motion.

• if the graph line is not straight, then it is non uniform motion.

• We can deduce different quantities and also derive equation of motion from graphs.

For example,

• from the distance – time graph, if we calculate the slope of the graph line, then we can find the velocity of the object.

• Also from the velocity – time graph, if we find the slope of the graph then we can find the acceleration of the object.

Uniform circular motion: When a body moves in a circular path with uniform speed (constant speed), its motion is called uniform circular motion.When a body moves in a circle, its direction changes continuously. Due to this change in direction we say that velocity changes and the motion is accelerated. From Newton’s 1^st law we know that to create acceleration an external force is required. Hence we need a force to keep the body moving in a circular path.This force is called Centripetal Force.

Uniformly Accelerated motion: When a body moves along a straight line and its velocity increases by equal amounts in equal interval of time then the motion is called uniformly accelerated motion.

E.g.:Motion of a freely falling body.

Let us consider an object moving in straight line with a constant acceleration. For such a situation the first equation of motion gives the final velocity (v) after time t given that the object is having an initial velocity of u and constant acceleration of a. Now for deriving it, let us consider the following velocity time graph:

Initial velocity of the body, u = OA ... (1)

Final velocity of the body, v = BC ... (2)

From the graph BC = BD + DC

Therefore, v = BD + DC ... (3)

Again DC = OA

So, v = BD + OA

Now from equation (1), OA = u

So, v = BD + u ... (4)

We should find the value of BD now

Now, from the velocity time graph, if we calculate the change in velocity to change in time, then we can find the acceleration, a. This is nothing but the slope of the graph.

Thus,

Acceleration, a = Slope of line AB

a =

But AC = OC = t

Hence we get

a =

BD = at

Now putting this in equation (4) we get

v = u + at

Uniform motion: When a car moving at a constant velocity of 50Km/hr in a straight line, then the car is covering equal distances in equal intervals of time irrespective of the length of time. Then we can say that the car is in uniform motion.

Non Uniform motion: When a race car constantly accelerates to win the race, then the car travels unequal distances in equal intervals of time. Hence such an instance can be referred as an example of non-uniform motion.

Uniform circular motion: When a table fan rotates at a constant speed then it can be described as an uniform circular motion as it is rotating in a circle in which at all instances change of direction is taking place and movement is also circular.

Non uniform circular motion: Motion of a table fan when its switched off till its stops, can be referred as an example of non-uniform circular motion as the rotation of the fan is not constant and it is changing with time.

Uniformly accelerated Motion: When a body falls freely, then it accelerates uniformly with a magnitude of acceleration due to gravity. Thus velocity of the body increases by equal amounts in equal intervals of time. Hence it can be referred as an example of uniformly accelerated motion.

Non-uniformly accelerated motion: When a feather falls down in heavy storm, then the velocity of the feather decreases sometimes and sometimes it increases in unequal amounts in equal intervals of time. Thus it can be referred as an example of non-uniformly accelerated motion.

Height = 20cm

Width = 20cm

Distance to climb 1 stair = 20 + 20 = 40cm

Distance to climb 5 stairs = AB + BC

= 5*40

= 200cm

Displacement =

=

Frequency = ʋ

ʋ = 100 rpm

= 100/60 s^-1

= 5/3 Hz(s^-1)

Angular velocity = ω

= 2*π*ʋ

= 2 × 3.14 × 5/3

= 10.47rad s^-1
r = 0.20m

Linear velocity v = ω*r

= 10.47 × 0.20

= 2.094ms^-1

We simply plot the points on the graph containing

X axis as Time

Y axis as Velocity

Velocity v-s time graph of given data is

Deduction: We can easily see that the change in velocity per unit time is not constant. In other words velocity time graph is not a straight line, hence the motion is Non Uniformly Accelerated motion.

Initial velocity = u

u = 6m/s

Final velocity = v

V = 16m/s

Time = t = 10 s

• Acceleration = a

a = (v-u) / t

= (16-6)/10

= 1 m/s²

• Distance = s

s =

s =

= 110 m

Circumference = 3140 m

Speed = 10 m/s

a) Distance of scooter = circumference/2

= 3140/2

= 1570 m

b) Displacement of scooter = AB

= circumference/ π

= 3140/3.14

= 1000 m

c) Time taken = distance/speed

= 1570/10

= 157 s

Acceleration: Acceleration of a body is defined as the rate of change of its velocity with time

Acceleration =

SI unit: ms^-2

For a body moving in a straight line we consider it to be moving with

a) positive acceleration if the speed increases with time

b) negative acceleration if speed decreases with time

An example of uniform acceleration is: a ball rolling down a frictionless inclined plane.

Whenever we have to find total displacement from a velocity time graph we find the area under the curve.

In the figure we clearly see

• Two triangles ABF & DCE having area A₁ & A₂ respectively

• A rectangles BEIF having area A₃.

A₁ = = 5

A₂ = = 10

A₃ = (5-0)*(10-2) = 40

Total area under the curve = A₁ + A₂ + A₃

= 5 + 10 + 40 = 55

Hence total displacement = 55 m

1. a) Body A moves with constant speed since it travels equal distance in equal interval of time.

b) Body B moves with constant acceleration since the rate of change of distance per second in it is constant.

2. a) Body C covers the maximum distance in 3^rd second.

C covers 100 cm whereas A & B cover 20 cm & 24 cm in 3^rd second.

b) Body A covers the minimum distance in 3^rd second.

Since A covers only 20 cm which is less than covered by B (24 cm) and C (100 cm)

3. Body C is moving with non-uniform acceleration.

Following is the graph corresponding to the uniformly accelerated motion

a) Velocity at 2.5 seconds is 7 m/s.

b) Acceleration = slope of velocity time graph

=

= 2 m/s²

c) Distance covered in last 4 seconds is the area under the curve for last 4 second

i.e.

Total distance covered in last 4 seconds

= area under ABCD

= area of ABED + area of BCE

= ((6-0)*(6-2)) +

= 24 + 16

= 40 m

(a) Mehak is curious, practical, skilled, alert and enthusiastic.

(b) Data given is insufficient to decide what kind of motion she infers.Motion can be uniform or non-uniform.

(c) Average speed =

And total distance travelled is given by the area under the velocity v/s time graph

Hence,

Average speed =