xam idea class 9 Science Chapter 8 Motion

Proficiency Exercise
1) Draw velocity–time graph, when an object has uniformly accelerated velocity.

2)A body thrown in the vertically upward direction rises upto a height 'h' and comes back to the position of its start. Calculate the displacement of the body.
Displacement = 0

3)State a relationship connecting u, v, a and t for an accelerated motion.
v = u + a t

4)What does the odometer of an automobile measure?
Odometer of an automobile measures distance travelled by the car.

5)What is the acceleration of a body moving with uniform velocity?
Zero (0)

6)Uniform circular motion is an accelerated motion. Justify this statement giving an example or activity.
Uniform circular motion is an accelerated motion because the magnitude of speed is constant but direction of velocity is changing constantly, e.g., a cyclist moving in a circular path with constant speed.

7)When is a body said to have

1. uniform velocity
2. variable velocity?

1. Uniform Velocity - Body has a uniform velocity if it travels in a straight line and covers equal distance in equal intervals of time.
2. Variable  Velocity -

A body is said to have variable velocity if 

1. speed of body changes.
2. When speed remains constant but direction of motion of body changes.

8) Can a body have acceleration without change in magnitude of velocity? Explain.

Yes, a body can have acceleration without change in magnitude of velocity, e.g., body moving in circular motion.

In this, the magnitude of velocity is same at every instant but the direction is changing constantly hence there is change in velocity & thus motion is said to be accelerated. 

9)An object travels 18 m in 4 s and next 18 m in 2 s. Calculate the average speed of the object.

AverageSpeed=TotaldistanceTotaltimetaken$$

Total distance = 18 + 18 = 36 m

Total time taken = 4 + 2 = 6 s

Average speed = 366$$ m/s = 6 m/s.

10) 200 m long train crosses a 400 m bridge with a velocity of 36 km/h. Find the time taken by the train to cross the bridge.

Here,

v = 36 km/h

As,1 km = 1000 m

∴ 1 h = 3600 s 

1 km/h = 10003600=518m/s$$

Hence, v = 36×518m/s=10m/s$$

Total distance = 200 + 400 = 600 m

Velocity = 36×518m/s=10m/s$$

⇒   Time =  DistanceVelocity=60010=60s=1minDistanceVelocity=60010=60s=1min

11)A scooter acquires a velocity of 36 km/h in 10 s just after the start. Calculate the acceleration of the scooter.

Here, u = 0

v = 36 km/h

=  36×518m/s=10m/s$$

t = 10 s

Using, v = u + at

10 = 0 + a × 10

a=1010=1ms–2

12)A racing car has a uniform acceleration of 4 ms–2. What distance will it cover in 10 seconds after the start?

Here, a = 4m/s2,  t = 10 s, u = 0 m/s

s = ?

Using, s = ut + 12$$at2 

= 0 × 10 + 12$$× 4 × 100= 200 m

13)Distinguish between uniform and non-uniform motion.

Uniform Motion

Body is said to be in uniform motion if it covers equal distance in equal intervals of time.

Example:  A car moving with  a uniform speed of 10 km/h.

Non-uniform Motion

Body is said to be in non-uniform motion if it covers unequal distance in equal intervals of time. 

Example:  Motion of a freely falling body.

 

14)A car travels from stop A to stop B with a speed of 30 km/h and then returns back to A with a speed of 50 km/h. Find :

1. Displacement of the car.
2. Distance travelled by the car.
3. Average speed of the car.

1. Displacement of the car is zero as it comes back to initial point.
2. Distance cannot be calculated as total time is not given.
3. t1=x30,t2=x50$$

Average speed = 2xx30+x50$$

= 2xx(130+150)$$

=25+3150=2×1508$$

=752$$km/h

15) An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement (a) When it completes half the circle (b) When it completes full circle (c) What type of motion does the object possess? (Justify your answer)

1. When it completes half circle,
   AB = 2 r is the displacement.
   Distance = πr     
2. When it completes full circle, it is back to the starting point & thus displacement is zero.
    Distance = 2 πr
3. The object possesses the accelerated motion. Since the velocity changes (due to continuous change in direction), therefore, the motion along a circular path is said to be accelerated.

16)Study the velocity–time graph of an object given below and answer the following questions:

1. Which part of the graph represents uniform positive acceleration? Calculate this acceleration.
2. Calculate the distance travelled by the object from A to B.

1. OA represents positive acceleration
   a =  v2–v1t2–t1=4–04–0=44$$
   a = 1 m/s2
2. Distance travelled by object from A to B=area of rectangle ABCD=AB × BD AB = 10 – 4 = 6 s
   BD = 4 m/s
   Distance = Area of ABCD= AB × BD
   = 6 × 4 = 24 m

17)The brakes applied to a car produce an acceleration of 6 m/s2 in the opposite direction to the motion. If the car takes two seconds to stop after the application of brakes, calculate the distance it travels during this time.

Here, a = –6 m/s2 (retardation)

t = 2 s

v = 0

Using v = u + at

u = – at = –(–6) × 2 = 12 m/s

v2 = u2 + 2 a s

s = –u22a$$

= –12×122×(–6)=12m

18)If the reading on the odometer of a vehicle in the beginning of a trip and after 40 minutes were 1048 km and 1096 km respectively, calculate its average speed. Will the reading on the speedometer show this speed when the vehicle is moving? Support your answer with reason.

Here, t1= 0 min,     t2 = 40 min

s1 = 1048 km,        s2 = 1096 km

As, 1 h = 60 min

1 min = 160$$h

t2 = 40 min =  4060h=23h$$

Average speed = TotaldistancetravelledTotaltime=s2–s1t2–t1$$

=  1096–104823=48×32$$

= 72 km/h

Speedometer will not show this reading because it shows speed at any instant which keeps on changing.

19)Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 ms–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 ms–1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases.

Here, vi = 6 m/s, u = 0

t = 30 s

After t = 30 s, brakes are applied

v = 4 m/s in t2 = 5 s

Using

v = u + a t 

a=v–ut=+630$$m/s2

= 15$$m/s2 = 0.2 m/s2

vf = 4 m/s, vi = 6 m/s, t = 5 s

a = vf–vit=4–65$$ = –0.4 m/s2

20)A ship is moving at a speed of 56 km/h. One second later, it is found to be moving at 58 km/h, what is the acceleration of the ship?

u = 56 km/h

t = 1 s   =160×60h=13600h$$

v = 58 km/h

Using a = v–ut$$

a=(58–56)km/h13600h$$

= 2 × 3600 km/h2

= 7200 km/h2

21)A train starting from rest attains a velocity of 72 kmh–1 in 5 minutes. Assuming that the acceleration is uniform, find

1. the acceleration, and
2. the distance travelled by the train for attaining this velocity.

1. Here,  u = 0, v = 72 km/h
   t = 5 minutes = 5 × 60 s
   = 300 s
   v = 72 km/h
   v = 72×518m/s=20m/s$$a =  v–ut=20–0300=115m/s2$$
2. Using, s = u t + 12$$a t2
   u = 0, a =   m/s2, t = 300 s
   s = 115$$
    = 12×115×(300)2$$
    300×30030$$ = 3000 m
   =3km

22)Write three equations of motion and derive them.

Three equations of motion

1. To Derive: v = u + a tAcceleration = changeinvelocitytime$$
   = Finalvelocity–Initialvelocitytime$$
   a=v–ut⇒v=u+at$$
2. To Derive: s = u t  +12at2$$
   Averagevelocity=Initialvelocity+finalvelocity2$$
   =u+v2$$
   s = Average velocity × time
   v  = u + a ts = vt2+ut2$$
   = ut+at22+ut2=ut+12at2$$
   s=ut+12at2$$
3. To Derive: v2 = u2 + 2 a ss = ut+12at2$$.    ... (1)
   v = u + a t ... (2)
   t=v–ua$$... (from (2))
   s=u(v–u)a+1a2(v–u)2a2$$... (Putting value of ‘t’ in (1)
   s=uv–u2a+12(v–u)2a$$
   2 a s = 2 u v – 2 u2 + v2 + u2 – 2 v u2 a s = v2 – u2

23)Account for the following:

1. Name the quantity which is measured by the area occupied below the velocity-time graph.
2. An object is moving in a certain direction with an acceleration in the perpendicular directions.
3. Under what condition is the magnitude of average velocity of an object equal to its average speed?
4. An example of uniformly accelerated motion.
5. A body is moving along a circular path of radius (R). What will be the distance and displacement of the body when it completes half revolution?

1. Area occupied by v – t graph gives distance.
2. It is an example of circular motion where the acceleration is towards the centre.
3. The magnitude of average velocity of object is same as average speed when the object is not changing its direction.
4. An object freely falling under gravity.
5. When a body moves half revolution
    Distance = π RDisplacement = AB = 2 R

24) 

Define circular motion.

1. What is the difference between uniform motion in a straight line and circular motion?
2. An athlete completes one round of a circular track of diameter 200 m in 40 s, what will be the distance covered and the displacement at the end of 2 min and 20 s? A ship is moving at a speed of 56 km/h. One second later, it is found to be moving at 58 km/h, what is the acceleration of the ship?

1. Circular motion - When a body moves in a circular path with uniform speed, the motion is uniform circular motion.
2. For uniform motion in a straight line, the speed of object and the direction in which the object moves is constant.
   Whereas in circular motion,
   Velocity of a body moving in circular path with constant speed is continuously changing because velocity = speed in a specified direction and here with change of direction, the velocity changes.
3. Diameter of track = 200 m
   Time = 40 s to complete one round of track.
   2 min 20 s=2 × 60 + 20
   =140 s=120 + 20
   =120 + 20 s = 40 × 3 + 20
   140 = (40) × 3.5
   So in 140 s, an athlete completes 3 complete rounds of circular track and one half circle.
    So, the Displacement for 3 complete rounds of circle and one half circle
   = 0 + Displacement due to half circle
   =2 r = Diameter
   =200 m
   Distance travelled by an athlete
   =3 × (2 π r) + π r=6 πr + πr = 7 πr= 7 × 3.14 × (D/2) = 7 × 3.14 × 100
   =2198 m

25)A circular cycle track has a circumference of 314 m with AB as one of its diameters. A cyclist travels from A to B along the circular path with a velocity of constant magnitude 15.7 m/s. Find:

1. the distance moved by the cyclist.
2. the displacement of the cyclist if AB represents north-south direction.
3. the average velocity of the cyclist.

2 πR = 314 m

u = 15.7 m/s

1. Distance moved by cyclist in moving from
    A to B = πR = 314/2 m = 157 m
2. Displacement = 2 R in North South
   π R = 157
   R = 157/3.14 = 50 m
   Displacement = 100 m
3. Magnitude of velocity = 15.7 m/s
   Time=DistanceSpeed=15715.7=10s$$
    Average velocity = DisplacementtravelledTimetaken$$
   = 10010$$
   = 10 m/s

26)What does displacement-time graph represent? How will you determine velocity of the body from this graph? 

Displacement-time graph can be used to calculate velocity of object.

Velocity of the body = d1–d0t1–t0$$

here in this graph, d0 = 0,  t0 = 0

v = d1t1$$  along A B

V B C  =  d1–d1t2–t1=0$$

The body is moving uniformly from time t1 to t2.

HOTS (Higher Order Thinking Skills)
1)
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure. Which car is the slowest?

Speed = Slope of distance–time graph. The smaller the slope, the smaller is the speed.

From the figure, slope is minimum for car D. So, D is the slowest car.

2)

A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in figure. Plot a velocity–time graph for the same.

 

Velocity from 0 to 50 s is v1 = ΔsΔt=100–050–0$$ = 2 ms–1

Velocity from 50 s to 100 s, v2 = ΔsΔt=0–100100–50$$ = –2 ms–1

Accordingly the velocity-time graph is shown in figure below.

3)

Suppose a squirrel is moving at a steady speed from the base of a tree towards some nuts. It then stays in the same position for a while, eating the nuts, before returning to the tree at the same speed. A graph can be plotted with distance on the x-axis and the time on y-axis.

Observe the graph carefully and answer the following questions.

1)

Which part of the graph shows the squirrel moving away from the tree?

Part AB

2)Name the point on the graph which is 6 m away from the base of the tree.

Point B

3)Which part of the graph shows that the squirrel is not moving?

Part BC

4)Which part of the graph shows that the squirrel is returning to the tree?

Part CD

5)Calculate the speed of the squirrel from the graph during its journey.

Total distance travelled = 6 m + 6 m = 12 m

Time = 11 s.

Speed = Time distanceTime=1211$$ = 1.09 m/s.

4)

The table given below shows distance (in cm) travelled by bodies A, B and C. Read this data carefully and answer the following questions.

Distance (in cm) covered by different bodies

Time in (s)	Body (A)	Body (B)	Body (C)
1st Second 2nd Second 3rd Second 4th Second 5th Second	20 20 20 20 20	20 36 24 30 48	20 60 100 140 180

1. Which of the bodies is moving with
   1. constant speed?
   2. constant acceleration?
   3. non-uniform acceleration?
2. Which of the bodies covers
1. maximum distance in 3rd second?
2. minimum distance in 3rd second?

Ans 

1. 1. Body A
   2. Body C
   3. Body B
2. 1. Body C. Total distance travelled = 100 – 60 = 40 cm
   2. Body B. Total distance travelled = 24 – 36 = (–) 12 cm The negative sign implies decceleration.