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Correct option is A) 30 m

The mass of a body is constant and it does not change by changing the position of the body.

Motion – The change in the position of an object with time is called as motion.

Types of Motion : 

1. Straight line motion 

2. Circular motion 

3. Rotational motion 

4. Vibratory motion 

5. Periodic motion

(a) Rotational motion

The motion of a spinning top is rotational motion while the motion of the bull of a bull – crusher is circular motion.

The type of motion which repeats itself after a certain period of time is called as periodic motion.

Examples: 

1. Motion of the pendulum of a clock 

2. Children swimming on a swing.

To measure the length of a curved line ‘AB’ we take a long thread. We take one end of the thread and put it on the point ‘A’ of the curved line. Now, we move the thread along the curved line and hold the other end of the thread when it reaches the point ‘B’ . Now, we stretch the thread along a meter ruler and measure it. This is the length of the curved line.

1. Stone falling from a height. 

2. Car running on a straight road.

Examples of straight line motion 

1. Motion of a car on a straight road. 

2. Trains moving on a straight track 

3. Children sliding on a slide

Periodic motion

No, they perform different types of motions. Straight line motion, periodic motion and circular motion.

1. Actual distance = \(\overline{AB}+\overline{BC}+\overline{CD}+\overline{DE}\)

= 3 + 4 + 5 + 3

Actual distance = 15 km

2. Displacement = \(\overline{AB}+\overline{BD}+\overline{DE}\)

= 3 + 3 + 3

Displacement = 9 km

3. Speed = \(\frac{Distance\,travelled}{Total\,time}\)

Distance = 15 km = 15 × 1000 = 15000 m

Time = 1 hr = 1 × 60 × 60 = 3600 sec.

s = \(\frac{15000}{3600}\) or s = \(\frac{15\,km}{1\,hour}\) = 15km/hour

= 4.16 m/sec. or 15 km/hour

4. Velocity = \(\frac{Distance\,travelled}{Total\,time}\)

Displacement = 9 km = 9 × 1000 = 9000 m

Time = 1 hr = 1 × 60 × 60 = 3600 sec

V = \(\frac{9000}{3600}\) or V = \(\frac{9\,km}{1\,hour}\) = 9 km/hour

= 2.5 m/sec. or 9 km/hour

5. Yes, this velocity can be called as average velocity.

i. constant

ii. zero

iii. Speed

iv. Velocity

a) Vector :The physical quantity which is specified with magnitude and direction is called a vector. 

Eg : Displacement, velocity are vectors. 

b) Scalar : The physical quantity which does not require any direction for its specification is called ‘scalar’. 

Eg : Distance, time are scalars.

a) From the graph we can infer that in the interval OA, it’s almost a straight line with a positive slope. Therefore velocity is positive that is with increase in time distance also increases and thus acceleration is also positive as well as uniform. 

b) In the interval AB, the line has a negative slope as with increase in time distance decreases. Here velocity is negative and it shows retardation. 

c) BC represents more downfalls in the slope of the line which means velocity will become more negative and hence it shows retardation with greater magnitude than AB. 

d) CD shows that with increase in time distance is increasing again with a positive slope. Therefore velocity is positive which implies acceleration is also positive.

Using the first equation of motion: v = u + at

where: v = final velocity = 0 m/s

u = initial velocity = 0.5 m/s 

a = acceleration = -0.105 m/s² 

t = time = ?? 

v = u + at 

0 = 0.5 - 0.105 x t0.105 x t = 0.5 

t =\(\frac{0.5}{0.105}\) s =4.76 s

Initial velocity, u = 0 m/s 

Final velocity, v = ? Acceleration, a = 0.2 m/s² 

Time, t = 5 min = 300 s 

v = u + at = 60 m/s

The correct answer is d) 28 m/s

(i) Acceleration from A to B = \(\frac{change\,in\,velocity}{change\,in\,time}\)= \(\frac{25-0}{3-0}\)= 8.33 m/s² 

(ii) Acceleration from B to C = \(\frac{15-25}{4-3}\)= -10 m/s² 

(iii) Distance covered in the region ABE = \(\frac12\)× base × height 

= × 3s × 25 m/s = 37.5 m 

(iv) Average velocity from C to D = \(\frac{initial\,velocity\,+\,final\,velocity}{2}\)

= \(\frac{15-0}{2}\)= 7.5 m/s

(v) Distance covered in the region BCFE =\(\frac12\) × sum of parallel sides x 

distance between them =\(\frac12\) × ( 25+15) × 1 = 20 m

The correct answer is b) m²

i. False. 

C.G.S. unit of acceleration is cm/s².

ii. False. 

M.K.S. unit of force is Newton

iii. True

i. False. 

Speed is distance travelled per unit of time

ii. True

iii. True

iv. False. 

Change in velocity per second is acceleration

v. True

Distance travelled = A → B → C → D → E → F → G → H + I

= 5 + 7 + 6 + 3 + 5 + 4 + 6 + 5

= 41 m

Displacement = A → I in a straight line shortest distance

= 1m

The correct answer is c) motion of a racing car on a circular track

The correct answer is c) RS

Change in velocity = 0 - 100 km/h. = -100 km/h

= -100 \(\times\,\frac{5}{18}=\frac{-500}{18}m/sec.\)

Time =10 sec

Deceleration = \(\frac{{\frac{-500}{18}}}{10}=\frac{-500}{18}\,\times\,\frac{1}{10}=\frac{-50}{18}= \,-2.77\,m/sec^2\)

The correct answer is d) the distance travelled by the scooter is zero in this section

Force (F) = 2 N

Displacement (s) = 50 m

Work done (W) = ?

W = Fs 

= 2 × 50 

W = 100 Joule

The total distance the bird has traversed is the length of the distance covered by circling, but the eventual displacement are the bird is zero as its initial and final position are one and the same.

B) (a) = non – uniform; (b) = uniform

C) constant direction

A) acceleration

Correct option is B) 5 m

A) moving train comes to rest