Class 9 LAKHMIR SINGH AND MANJIT KAUR Solutions Physics Chapter 1 - Motion

No, displacement is a vector quantity.

Distance is a scalar quantity.

6 m/s

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

Speed of a body in a specified direction is called velocity.

(a) Motion of a bus on a road

(b) Motion of a racing horse

Speed is defined as the distance travelled per unit time.

(a) The speedometer of a car measures instantaneous speed of the car.

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

Speed gives an idea of how slow or fast a body is moving.

When the body comes back to its starting point, it has zero resultant displacement but covers a certain non-zero distance.

In addition to speed, we should know the direction in which the body is moving.

When a body covers equal distances in equal intervals of time in a particular direction however small or big the time interval may be, the object is said to have uniform velocity.

When the object moves in a single straight line, the magnitude of average velocity equal to average speed.

Average velocity of a moving body can be zero.

Motion of a boy from his home to shop (in one direction) and back to home (in its reverse direction) is an example of a situation in which a body has a certain average speed but its average velocity is zero.

When a body is moving with uniform velocity, its acceleration is zero.

Negative acceleration is also called retardation.

(a) Speed (or Velocity)

(b) Acceleration

Uniformly accelerated motion

S.I. unit of retardation is m/s².

(a) vector, scalar

(b) velocity

(c) acceleration, second, velocity, 3m/s

(d) displacement, m/s

(e) velocity, m/s²

A freely falling body has non-uniform motion because it covers smaller distances in the initial '1 second' intervals and larger distances in the later '1 second' intervals, i.e., it covers unequal distances in equal intervals of time.

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

For bus X,

Speed= Distance/Time

Speed=360/5=72km/h

For bus Y,

Speed= Distance/Time

Speed=476/7=68 km/h

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

Speed of athelete = 10 m/s

Speed of bicycle = 200 m/min = 200/60 m/s = 3.33 m/s

Speed of scooter = 30 km/h = 30000/3600 m/s = 8.33 m/s

3.33 m/s < 8.33 m/s <10 m/s

i.e. 200 m/min < 30 km/h < 10 m/s

(a)Acceleration=

(b) u = 0 m/s

v = 21 m/s

Time, t = 1 min = 60 sec

(a) Acceleration

(b) Retardation

(c) No, because if a body takes a round trip such that its final position is same as the starting position, then the displacement of the body is zero but the distance travelled is non-zero.

Average speed = Total distance travelled/ Total time taken

Total distance travelled = 100m = 0.1 km; Total time taken = 50 hr

Average speed= 0.1/50=0.002km/h

Total distance=100m =0.1 km

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

Average speed = Total distance travelled/ Total time taken

                       =0.1/0.25= 0.4km/h

Total distance travelled=100m

Total time taken = 9.83 sec

Average speed = Total distance travelled/ Total time taken

                     =100/9.83 =10.172m/s

Averge speed in km/h:

10.172 x(3600/1000)=36.62 km/h

Initial velocity= 0m/s

Final velocity=6m/s

Time=3 sec

Acceleration=

=

Initial velocity, u = 600 km/h

Final velocity, v = 1100 km/h

Acceleration = 10 km/h/s = 600 km/h²

From relation, a = (v-u)/t

t = (v-u)/a

t = (1100-600)/600 = 500/600 = 5/6 hr = 50 sec

Deceleration, a=-5m/s²

Initial velocity, u=20m/s

Final velocity, v=0m/s

t=?

(a) Distance travelled is the actual length of the indirect path covered by the body whereas displacement refers to the straight line path between the initial and final positions. For e.g. In the figure given below, a body moves from point A to point B and then from point B to point C. Here, the distance travelled by the body is AB + BC and displacement is AC.

(b)

PQ=8cm

QR=6 cm

Resultant Displacement PR =

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

A body has uniform motion if it travels equal distances in equal intervals of time, no matter how small these time intervals may be. For example: a car running at a constant speed of 10m/s, will cover equal distance of 10m every second, so its motion will be uniform.

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 .

(a) Speed of a body is the distance travelled by it per unit time. The SI unit of speed is m/s.

(b) (i) Average speed of a body is the total distance travelled divided by the total time taken to cover this distance.

(ii) Uniform speed refers to the constant speed of a moving body. A body has a uniform speed if it travels equal distance in equal intervals of time, no matter how small these time intervals may be.

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

( b)       (i)Speed is a scalar quantity whereas velocity is a vector quantity.

(ii) Speed of a body is distance travelled by it per unit time whereas velocity of a body is the distance travelled by it per unit time in a given direction.

(iii)Speed is always positive whereas velocity can be both positive as well as negative.

    (c) Speed = 54km/h = 54 x (1000/3600) = 15m/s

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

(b) A body has uniform acceleration if it travels in a straight line and its velocity increases by equal amounts in equal intervals of time. For example: Motion of a freely falling body.

Distance travelled in half a rotation of a circular path is equal to the circumference of semi-circle, i.e., =R.

Displacement= diameter of circle= 2R

(i) Distance travelled = 6 km

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

(i) Total distance travelled= 3 + 4 +9=16 km

(ii) The body travels a total distance of 12 km in east direction i.e. towards x-axis.

And it travels a distance of 4 km in North direction, i.e. towards y-axis.

Hence, resultant displacement is

=

(a) Total distance covered in going to the bookshop and coming back to the classroom = 20 + 20 = 40m

           Total time taken= 25 + 25 = 50 sec

  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 km

Initial velocity, u=6m/s

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

Time, t = 0.040 s

The motion is accelerated.

It represents uniform velocity.

Distance travelled by the moving body .

The slope of a speed-time graph indicates acceleration.

The slope of a distance-time graph indicates speed.

Motion of moon around the earth.

Uniform circular motion.

The Speed of the body is constant or uniform.

(a) Speed

(b) Direction of motion

The body has uniform speed.

The body is not moving. It is stationary.

It represents non-uniform acceleration.

It is accelerated motion as the velocity is changing continuously.

The tip of the 'seconds' hand' of a watch represents uniform circular motion. It is an accelerated motion.

(a) zero

(b) speed

(c) acceleration

(d) distance travelled

(e) circular path

Yes, uniform circular motion is accelerated because the velocity changes due to continuous change in the direction of motion.

The speed of a body moving along a circular path is given by the formula:

where, v= speed

 =3.14 ( it is a constant)

t= time taken for one round of circular path

The motion of a body which is moving with constant speed in a circular path is said to be accelerated because its velocity changes continuously due to the continuous change in the direction of motion.

Uniform linear motion is uniform motion along a linear path or a straight line. The direction of motion is fixed. So, it is not accelerated. For e.g.: a car running with uniform speed of 10km/hr on a straight road.

Uniform circular motion is uniform motion along a circular path. The direction of motion changes continuously. So, it is accelerated. For e.g.: motion of earth around the sun.

An important characteristic of uniform circular motion is that the direction of motion in it changes continuously with time, so it is accelerated.

Centripetal force brings about uniform circular motion.

Initial velocity, u=?

Final velocity, v=0m/s (car is stopped)

Retardation, a=-2.5 m/s²

Time, t=10s

v=u + at

0=u +(-2.5)x 10

u=25m/s

The velocity of this body is increasing at a rate of '10 metres per second' every second.

Initial velocity, u=0m/s

Time, t=2s

Acceleration, a=10m/s²

Using ,

Initial velocity, u=5m/s

Final velocity, v=?

Acceleration, a=0.2m/s²

Time, t=10 sec

Using , v=u + at

v=5 + 0.2 x 10

v=5 + 2 =7 m/s

Now distance travelled in time is calculated;

Using,

Initial velocity, u=18km/h

Final velocity, v=0m/s

Time, t=2.5 sec

Acceleration, a=?

Using , v= u + at

So, retardation is 2m/s².

Initial velocity, u=0m/s

Final velocity, v=?

Acceleration, a=0.2 m/s²

Time, t=5min= 5 x60=300 sec

Using, v = u + at

v =0 + 0.2 x 300=60m/s

And the distance travelled is

(a) Distance and Time

(b) Speed (or velocity) and Time

Initial velocity, u=0m/s

Final velocity, v=?

Acceleration, a=2m/s²

Time, t=10s

(a) Using, v = u + at

v=0 + 2 x 10= 20 m/s

(b) Distance travelled is:

Initial velocity, u=20m/s

Time, t=30 s

Acceleration, a=0.5m/s²

Distance travelled is:

Initial velocity, u=15m/s

Final velocity, v=0m/s

Distance, s=18m

Acceleration, a=?

So, deceleration is 6.25 m/s².

(a) The train has a uniform velocity.

(b) There is no acceleration.

(a) v=u + at is the first equation of motion. It gives the velocity acquired by a body in time t when the body has initial velocity u and uniform acceleration a.

(b) Initial velocity, u=0m/s

Time, t=5 s

            Distance, s=100m

            Acceleration, a=?

(a) Consider a body having initial velocity 'u'. Suppose it is subjected to a uniform acceleration 'a' so that after time't' its final velocity becomes 'v'. Now, from the definition of acceleration we know that:

(b) Initial velocity, u=54km/h= 15m/s

Final velocity, v=0m/s

Time, t=8s

          Acceleration, a=?

(a) Suppose a body has an initial velocity 'u' and a uniform acceleration' a' for time 't' so that its final velocity becomes 'v'. Let the distance travelled by the body in this time be 's'. The distance travelled by a moving body in time 't' can be found out by considering its average velocity. Since the initial velocity of the body is 'u' and its final velocity is 'v', the average velocity is given by:

(b) Initial velocity, u=0m/s

Final velocity, v=36km/h=10m/s

Time, t=10min =10 x 60=600 sec

          Acceleration, a=?

(a) When a body moves in a circular path with uniform speed (constant speed), its motion is called uniform circular motion. For e.g.

(i) Artificial satellites move in uniform circular motion around the earth.

(ii) Motion of a cyclist on a circular track.

(b) The speed of a body moving along a circular path is given by the formula:

Given, t=60 sec

Radius, r=10.5cm=0.105 m

Consider the velocity-time graph of a body shown in figure.

The body has an initial velocity u at a 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 the slope of a velocity-time graph is equal to the 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:

or BD=at

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

v= u+ at

Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a 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.

Suppose the body travels a distance s in time t. In the 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

Now, we will find out the area of rectangle OADC and area of triangle ABD.

(i) Area of rectangle OADC =OA x OC

= u x t

                   =ut

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

=_(1/2) x AD x BD

=_(1/2) x t x at

=_(1/2)at²

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

Consider the velocity-time graph of a body shown in figure. The body has an initial velocity u at a 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.

The distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

Distance travelled, s= Area of trapezium OABC

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

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

(i) The distance covered from A to B is ( 3-0) =3 cm

          Time taken to cover the distance from A to B =(5 -2) =3s

(ii) The speed of the body as it moves from B to C is zero.

     (iii)  The distance covered from C to D is (7-3)=4 cm 

          Time taken to cover the distance from C to D = (9-7)=2s 

(a) The body has a uniform velocity if its displacement-time graph is a straight line.

(b) The body has a uniform acceleration if its velocity-time graph is a straight line.

(i) BC represents uniform velocity. From graph, we see that the velocity of the body at point C = 40km/h

(ii) Acceleration between A and B = slope of line AB

=

(iii) BC represents uniform velocity, so acceleration acting on the body between B and C is zero.

Distance travelled = Area of rectangle OABC

= OA x OC

= 5 x 5 =25 m

(i) Initial speed of the car=10km/h

(ii) Maximum speed attained by the car= 35km/h

(iii) BC represents zero acceleration.

(iv) CD represents varying retardation.

(v)

(i) Graph (c): The speedof the ball goes on decreasing uniformly as it moves upward, reaches zero at the highest point, and then increases uniformly as it moves downward.

(ii) Grap(a): The speed of the trolley decreases uniformly, then it moves at a constant speed, and then the speed increases uniformly.

(i) OA represents uniform acceleration

(ii) AB represents constant speed.

(iii) BC represents uniform retardation.

(iv) Acceleration of car from O to A = slope of line OA

(v) Acceleration of car from A to B is zero as it has uniform speed during this time.

(vi) Retardation of car from B to C = slope of line BC

(i) Graph (a) represents uniform acceleration.

(ii) Graph (b) represents constant speed.

(iii) Graph (c) represents uniform retardation.

(iv) Graph (d) represents non-uniform retardation.

Initial velocity, u=8m/s

Acceleration, a=1m/s²

Distance, s=18m

Initial velocity, u=20m/s

Final velocity, v=0m/s

Distance, s=50m

The car's deceleration must be 4 m/s².