Motion

distance is a scalar quantity since it does not depend on direction.

Speed = 6m/s

To convert m into Km we divide it by 1000 and to convert s into hr we divide it by 3600

= (6 x 3600)/1000 = 21.6 Km/hr

Velocity is a vector quantity hence it gives speed in a specified direction.

Two examples of bodies having non-uniform motion is

a. motion of car on a road

b. motion of an aeroplane

Speed is the physical quantity obtained by dividing ‘Distance travelled’ by ‘Time Taken’ to travel that distance

(a) speedometer measures instantaneous speed of the car.

(b) odometer is a device used to record the distance travelled by the car.

Speed is the physical quantity obtained by dividing ‘Distance travelled’ by ‘Time Taken’ to travel that distance

A body travel a certain distance and yet its resultant displacement by zero when the body comes back to its starting point.

To predict the position of a moving body we should know the speed and direction of speed.

a body is said to have uniform velocity, when

1) Body covers equal distance in equal interval of time.

2) In particular direction

the magnitude of average velocity is equal to average speed when body moves along a straight line path.

Average velocity can be zero under certain conditions

A car going from home to the market and then back home.

Acceleration = velocity/ time = 0/ time = zero

The other name of negative acceleration is Deceleration.

a) velocity/ speed – m/s

b) Acceleration or retardation – m/s²

Uniformly accelerated motion is exhibited by a freely falling body.

Retardation is another name for acceleration and it is m/s²

a. Displacement, Distance

>Displacement depends on direction but distance does not.

b. speed, direction

>velocity = displacement/ time and is directional

c. speed, 3m/s

> acceleration = speed/time

d. Velocity

> Velocity is directional hence it is displacement/ time

e. Acceleration, m/s²

^>a = v/t

(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

(b) All of them never come at the same point at the same time.

(c) According to graph; each small division shows about 0.57 km.

A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km

Thus, at this point C travels about

9.14 – (0.57 x 3.75) km = 9.14 km – 2.1375 km = 7.0025 km ≈ 7 km

Thus, when A passes B, C travels about 7 km.

(d) B passes C at point Q at the distance axis which is ≈ 4km + 0.57km x 2.25 = 5.28 km

Therefore, B travelled about 5.28 km when passes to C

(a) Distance travelled by car in the 4 seconds

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second.

On the time axis, 5 squares = 2s

∴ 1 square =25s=25s

On the speed asix 3 squares = 2 m/s

∴ 1 square =23m/s=23m/s

∴ area of 1 square =25s×23m/s=415m=25s×23m/s=415m

∴ area of 62 squares =415m×62=415m×62

24815m=16.53m24815m=16.53m

Therefore, car travels 16.53 m in first 4 second.

(b) Part MN of the slope of the graph is straight line parallel to the time axis, thus this potion of graph represents uniform motion of car.

(i) Velocity of the body at point C is 40 km/h

(ii) Acceleration acting on the body between A and B is 6.6 km/h²

(iii) Acceleration acting on the body between B and C is zero

V = 5 m/s

T = 5 sec

On applying formula, we get

velocity= displacement / Time

Distance covered = 25 m

(i) the acceleration of the lift during the first two seconds is2.3 m/s²

(ii) the acceleration of the lift between second and tenth second is Zero

(iii) the acceleration of the lift during the last two seconds is2.3 m/s²

(i) the acceleration of the car is2 m/s²

(ii) the distance travelled by the car in 10 seconds is 140 m

(i) the initial speed of the car is10 km/h

(ii) the maximum speed attained by the car is 35 km/h

(iii) BC shows zero acceleration.

(iv) CD shows varying retardation.

(v) the distance travelled in first 8 hours is 242.5 km

(i) Graph c represents the case of a cricket ball thrown vertically upwards and returning to the hands of the thrower?

(ii) Graph a represents the case of a trolley decelerating to a constant speed and then accelerating uniformly?

(i) Uniform acceleration is represented by OA.

(ii) Constant speed is represented by AB.

(iii) Uniform retardation is represented by BC.

(iv) the acceleration of car from O to A is 4 m/s²

(v) the acceleration of car A to B is Zero

(vi) the retardation of car from B to C is 2 m/s².

a. The motion represented by first graph is Uniform acceleration

b. The motion represented by first graph is Constant speed

c. The motion represented by first graph is Uniform retardation

d. The motion represented by first graph is Non-uniform retardation

U = 8 m s^-1

a= 1 m s^-2

d = 18 m

On using the formua we get

V = 10 m/s^-1

Given:

U = 20m/s

s = 50 m

V = o m/s

On applying the formula, we get

A freely falling body exhibits a non-uniform motion because it covers unequal distances in equal intervals of time.

Speed is a scalar quantity as it has no direction but only magnitude.

For bus X,

Speed= Distance/Time

Speed=360/6=60km/h

For bus Y,

Speed= Distance/Time

Speed=427/7=61 km/h

Speed of bus X is less than that of bus Y. Hence, bus Y travels faster.

Speed of athelete = 20 m/s

Speed of bicycle = 400 m/min = 400/60 m/s = 6.66 m/s

Speed of scooter = 60 km/h = 60000/3600 m/s = 16.66 m/s

6.66 m/s < 16.66 m/s <20 m/s

i.e. 400 m/min < 60 km/h < 20 m/s

Acceleration=

Where v = final velocity

U = initial velocity

T = time taken

Given: v = 30 m/s

u = 0 m/s

a = ?

Formula used: a = (v-u)/t

a = (30 m/s - 0)/60 sec

a = (30 m/s)/60s

a = 0.5m/s²

Therefore, the acceleration is 0.5m/s²

Acceleration is the term used to denote the change of velocity with time

Retardation means the same as ‘moving with a negative acceleration

the distance travelled by the object will not be zero asdistance is a scalar quantity having magnitude only and no specified direction.

According to Question we have:-

Total Distance= 100 metres

We know that 1000 metres= 1 km ∴ 1 metres=

∴ 100 metres=

Average speed =

=

Given:

Total distance=1000 m

Total time taken=15 minutes= 15/60=0.25 hour

We know

1 km=1000 m

Total distance traveled=1 km

1 hour=60 minutes
15 minutes=0.25 hours

Total time taken=0.25 hours

putting the values of distance and time in the equation:



hence, the speed of tortoise is 4 km/h

Total distance travelled=1000m

Total time taken = 9.83 sec

Average speed = Total distance travelled/ Total time taken

=1000/9.83 =101.72m/s

Averge speed in km/h:

101.72 x(3600/1000)=366.2 km/h

Given:

speed from a to b=30 km/h
speed from b to a=20 km/h

Initial velocity= 0m/s

Final velocity=12 m/s

Time= 6 sec

Acceleration =

According to the Question,We have:-

Initial velocity, u = 600 km/h (Given)

Final velocity, v = 1100 km/h(Given)

Acceleration = 10 km/h/s

From relation, v = u+at,

On rearranging the terms, we get, v - u = at

or

∴ t = 50 seconds

Hence it will take the aircraft 50 seconds to reach the "sound barrier"

Deceleration, a=-5m/s2

Initial velocity, u=40m/s

Final velocity, v=0m/s

t=?

Distance travelled is the actual length of the path covered by the body whereas displacement refers to the straight line path from the initial to the final positions.

A body goes from point A to B and then comes back to point A.

Distance = 200 m

Displacement = 0 m

PQ=3cm

QR=4 cm

Resultant Displacement PR = sqrt (9 + 16)= 5 cm

when the position of a body changes continuously with respect to a stationary object taken as reference point, the body is said to be in motion.

A body has uniform motion if it travels equal distances in equal intervals of time irrespective of the interval. Example: motion of earth around sun.

Non-uniform motion: A body has a non-uniform motion if it travels unequal distances in equal intervals of time. For example: dropping a ball from the roof of a tall building .

the distance travelled by a body per unit time is called speed. The SI unit of speed is m/s.

(i) the total distance travelled divided by the total time taken to cover the distance by a body is called average speed.

(ii) A body has a uniform speed if it travels equal distance in equal intervals of time, irrespective of the interval of time.

the distance travelled by a body per unit time in a given direction is called velocity. SI unit of velocity is m/s.

(i)

Acceleration of a body is defined as the rate of change of its velocity with time. SI unit of acceleration is m/s².

Acceleration = Rate of change of velocity/ time

uniform acceleration is the motion of a body in a straight line whose velocity increases by equal amounts in equal intervals of time. For example: Motion of a freely falling body.

total distance =400 km
average speed = 70 km/h
time taken = 400/70=40/7 h

1)During the first 200 km

speed= 100 km/h
time taken t₁ =200/100=2 h

2)During the second 200 km

t₁ + t_{2 =} 40/7

2 + t_{2 =} 40/7

t₂₌ 26/7 h

speed = distance / time= 200×7/26=1400/26 =53.84 km/h

The speed of train during second 200 Km is 53.84 km/h to maintain an average speed of 70 km/h

(a) For the first part
In the first part, the train travels 15 km at a speed of 30 km h^-1
The speed of the train is given by
We know the speed of train =30 km/h
The distance travelled by tain= 15 km


Time taken by train in the first part, T₁=15/30=0.5 h

(b) For the second part
The distance, D= 75 km
The speed, S=50 km/h
Putting the values in the formula to calculate time taken
Time taken, T₂=75/50=1.5 h

(c) for the third part
The distance, D=10 km
The speed, S=20 km/h
Putting the values in the formula to calculate time taken
Time taken by train, ^{T₃=10/20=0.5} h.


Total distance travelled by train= 15 +75+10 =100 km
Total time taken by train=T₁+T₂+T₃ =0.5 +1.5+0.5=2.5 h

Putting the values in the above formula, we get
Average speed, S=100/2.5=40 km h^-1

Displacement is directional.

displacement to distance for a moving object is equal or less than 1

constant speed gives accelerated motion.

the distance moved and the magnitude of displacement are equal if the car is moving on straight road

0.12m/s x 3600/1000 = 0.432 Km/hr

increase in velocity is with increase in time.

Velocity is vector quantity.

velocity is a vector quantity.

velocity and displacement are directional.

a = v/t

speed = distance /time

displacement , acceleration and velocity are vector quantity.

a = v/t = distance / s²

When the body is completing half a revolution of a circular path of radius r then the:-

Distance traveled in half a rotation of a circular path is equal to the circumference of semi-circle=

Displacement is known as the shortest distance between two points,
as we can see the figure the shortest distance between point A and point B is along the diameter of the circle thus displacement is 2r

(i) Distance travelled = 12 km

(ii) Displacement = zero (since final position is same as initial position)

1) Total distance =3+ 4 + 9 = 16 Km

2) Total displacement = 12.6 km

The net displacement in east direction is 3+9=12 km

The net displacement in north direction is 4 km

The net displacement in south and west direction is zero

Using, pythagorean theorem

displacement(d)=√(12²+ 4²)=√(144+16)=√(160)=12.6 km

Hence, The displacement of body is 12.6 km in north-east direction

(a) Total distance covered in going to the bookshop and coming back to the classroom = 40 + 40 = 80m

Total time taken= 50 + 50 = 100sec

80/100 m/s = 0.8 m/s

0/100 m/s = 0 m/s

In the first case, car travels at a speed of 60 km/h for a distance of 100 km

In the second case, car travels at a speed of 40 km/h for a distance of 100 km

Total distance travelled = 200
Total Time Taken= t₁+t₂

=

According to the Question:-

Initial velocity, u=6m/s

Final velocity ,v=(-4.4m/s) (the ball rebounds in opposite direction)

∴

Speed remains constant in uniform circular motion.

Direction (of motion) change continuously in uniform circular motion.

True

the motion is accelerated when a body goes round the sun with constant speed in a circular orbit

It shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.

the quantity which is measured by the area occupied under the velocity-time graph is distance travelled.

the slope of a speed-time graph indicate acceleration.

the slope of a distance-time graph indicates speed.

The example of a motion where an object does not change its speed but its direction of motion changes continuously is the motion of moon around the earth

the type of motion in which a body has a constant speed but not constant velocity is Uniform circular motion.

When the motion of a body if its speed-time is a straight line parallel to the time axis the speed of body is constant.

It shows the graph for an object moving at a constant velocity. You can see that the displacement is increasing as time goes on. The gradient, however, stays constant (remember: its the slope of a straight line), so the velocity is constant. Here the gradient is positive, so the object is moving in the direction defined as positive.

It shows the graph for an object stationary over a period of time. The gradient is zero, so the object has zero velocity.

It shows the graph for an object moving at a constant acceleration. You can see that both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.

The motion is accelerated.

Accelerated motion is represented by the tip of ‘seconds’ hand’ of watch? Is it uniform or accelerated

(a) uniform velocity

(b) slope

(c) speed-time

(d) distance travelled

(e) circular path

Yes as the velocity changes due to continuous change in the direction of motion

the formula to calculate the speed of a body moving along a circular path is

⇒ speed of a body = (2 x constant x radius of the circular path) / (time taken for one round of circular path)

Due to the continuous change in the velocity there is a continuous change in the direction of motion because of which the motion of a body which is moving with constant speed in a circular path is said to be accelerated.

An important characteristic of uniform circular motion is that

a. The direction of motion in it changes continuously with time.

b. It is accelerated.

The force which brings about uniform circular motion is Centripetal force

t = 10 sec

a = -2.5 m/ s²

v = 0 m/s

v-u = at

On applying the formula, we get

u = 25m/s

a = 10m/s²

T = 10 sec

On calculating we get, that the velocity of this body is increasing at a rate of 10 metres per second.

On applying the formula, we get d = 20m

According to the Question,

u = 5m/s

a = 0.2 m/s²

We know that v= u+at, Putting the values in the equation,we get:-

v= 5+(0.2)x10

∴ v= 7 m/s

We know that v²-u²= 2as, Putting the values in the equation, we get:-

(7)²-(5)²= 2x(0.2)x s

49-25= 0.4s

∴ s =

u (initial velocity) = 18 Km/hr

t ( time )= 2.5 sec

v( final velocity) = 0 m/s

A = 0.2 m/s²(Given)

U = 0 m/s(As the train is starting from rest)

On applying the formula v= u+at we get

On applying the formula v² -u²= 2as we get

Given:

a is the acceleration of the cheetah = 2m/s²

t is the time period for which cheetah is accelerating = 10 sec

u is the initial velocity of the cheetah = 0 m/s

where
v is the final velocity of the cheetah
u is the initial velocity of the cheetah
a is the acceleration of the cheetah
t is the time period

=-Using the formula, we calculate distance travelled as 100m.

Given:
u = 20 m/s
a = 0.5 m/s²
t = 30 s
d= __ m

Formula:

using, Second equation of motion



d = 600 + 225 = 825 m

According to the Question we have :-

u = 15m/s (Given)

v = 0 (As she is applying brakes and coming to stop)

Using the Formula v²- u²= 2as

Putting the values we get:-

(0)²-(15)²= 2xax18

-225= 36xa

∴ a=

a. it has uniform velocity

b. it has no acceleration. As v/t = 0

final velocity is equal to sum of the initial velocity and product of acceleration and time.

According to Question we have :-

s=100m(Given)

t = 5 sec

u=0 (As the body is starting from rest)

We know that :-

Putting the values in the above equation we get:

Consider a body of mass “m” having initial velocity “u”.Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”.

Now we know that:
Acceleration = change in velocity/Time taken

=> Acceleration = Final velocity-Initial velocity / time taken

=> a = v-u /t
=>at = v-u

or v = u + at
This is the first equation of motion.

According to the Question we know:-

u = 54 Km/hr(Given)

(We know that 1km=1000m and 1hr= 60x60 seconds)

u=

t = 8 sec(Given)

^{0= 15+8a
⇒ (-15)= 8a}
∴ a= (-1.875 ms^-2)

Let the distance traveled by the body be “s”.

We know that, Distance = Average velocity × Time

Also, Average velocity =

.: Distance (s) = …….eq.(1)

Again we know that: v = u + at substituting this value of “v” in eq.(1), we get

s =

s =

s =

s =

or

This is the 2nd equation of motion.

u= 0m/s

V = 36 km per hour = 36 x (1000/3600) m/s

T = 10 minutes = 10 x 60 sec

On applying the formula,

Acceleration = 0.016 m/s²

v – u = at

s = ut +1/2 at²

v² = u² +2as

where,

u = initial velocity

v = final velocity

a = acceleration

t – time taken

s = distance travelled

Given:
u is the initial velocity = 0 m/s

v is the final velocity = 36 Km/hr = 36 x (1000/3600)=10 m/s

t is the time period = 10 ×60 =600 s

a = 2 m/s²

avg. velocity=

formula used:

1. final velocity " V "= u + a × t

where ,
v is the final velocity
u is the initial velocity
a is the acceleration of the body
t is the time period

2. Using third equation of motion

v² - u² =2 × a × s

where,
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement

putting the value of v,u,t in equation 1

10 =0 +a × 600

a = 0.0167 ms^-2

putting the value of v,u,a,t in equation 2

100-0=2×0.0167×s

s =3000 m= 3 km

When an object is experiencing uniform circular motion, it is traveling in a circular path at a constant speed. If r is the radius of the path, and we define the period, T, as the time it takes to make a complete circle, then the speed is given by the circumference over the period.

1. Twirling an object tied to a rope in a horizontal circle.

2. Revolution of earth

3. Hands of clock

t = 60 sec

r = 10.5 cm

v = 2π r/T

= 0.011 m/s

The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE.

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

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

But 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 out the value of BD now. We know that the slope of a velocity – time graph is equal to acceleration, a.

Thus, Acceleration, a = slope of line AB

or a = BD/AD

But AD = OC = t,

so putting t in place of AD in the above relation, we get:

a = BD/t

or BD = at

Now, putting this value of BD in equation (4) we get :

v = at + u

This equation can be rearranged to give:

v = u + at

And this is the first equation of motion. It has been derived here by the graphical method.

Velocity–Time graph to derive the equations of motion.

Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus:

Distance travelled = Area of figure OABC

= Area of rectangle OADC + Area of triangle ABD

We will now find out the area of the rectangle OADC and the area of the triangle ABD.

(i) Area of rectangle OADC = OA × OC

= u × t

= ut ...... (5)

(ii) Area of triangle ABD = (1/2) × Area of rectangle AEBD

= (1/2) × AD × BD

= (1/2) × t × at (because AD = t and BD = at)

= (1/2) at²...... (6)

So, Distance travelled, s = Area of rectangle OADC + Area of triangle ABD

or s = ut + (1/2) at²
This is the second equation of motion. It has been derived here by the graphical method.

Velocity–Time graph to derive the equations of motion.

We have just seen that the distance travelled s by a body in time t is given by the area of the figure

OABC which is a trapezium. In other words,

Distance travelled, s = Area of trapezium OABC

Now, OA + CB = u + v and OC = t. Putting these values in the above relation, we get:

...... (7)

We now want to eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion. Thus, v = u + at (First equation of motion)
And, at = v – u or

Now, putting this value of t in equation (7) above, we get:

or 2as = v² – u²[because (v + u) × (v – u) = v² – u²]

or v² = u² + 2as

This is the third equation of motion.

Let's put kinematics to play, shall we?

Consider the third equation of motion,

v²−u² =2×a×s ------- (1)

Here , 'v' is the final velocity of the body in question

' u ' is the initial velocity.

' a ' is the acceleration experienced.

' s ' is the distance traveled.

Taking only S.I. units, for the case mentioned in the question,

it can be calculated that a = -1/8 ms^-2 (v = 0, u = 10, s = 400).

The negative sign indicates that the body (the car) is decelerating.
Applying this result to first equation of motion,

v=u + a×t -------- (2)

You can calculate the time taken (t) as t = -10/(-1/8).

Which is:t = 80 seconds.

Speed of a car is not constant.

Given:
The initial velocity u =36 km/h=10 m/s
the Final velocity v =72 Km/h = 20 m/s
The time period t = 10 s

using first equation of motion
v – u = at
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time period

putting the value of v,u,t in the equation, we get

20 -10 =a ×10

a = 1 m/s²

V – u = at

a = v/t

a = v/t

when displacement is proportional to the square of time then object moves with uniform acceleration.

Speed is directly proportional to the slope of distance time graph

Since the velocity time graph for this car is a horizontal line parallal to the time axis then the velocity of the car will be constant. So, velocity of car at the end of 25 s will be 10m/s.

Velocity is directional.

.

Speed = constant

So distance = 0