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The length of pace or a footsteps various from person to person. If the length of foot- steps will be the used as standard unit for measurement of length then two measured quantity will not be same, Hence footsteps or pace is not a constant quantity and cannot be used as a standard unit of length.

On land – Buses and Trains 

On water – Ships and Boats 

On air – Helicopters and Aeroplanes.

i. Motion of a child on a swing is Periodic motion 

ii. Motion of the needle of a sewing machine is Periodic motion 

iii. Motion of wheel of a bicycle is Circular motion

(a) 1.65 m, as one metre = 100 cm = 1.65 x 100 cm =165 cm

(b) 1.65 x 100 x 10 mm = 1650 mm.

• Oscillations of a pendulum
• The motion of swing/motion of Earth around the sun

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

1. 100

2. 5000

3. periodic (oscillatory) motion

4. periodic oscillatory

5. circular

a. Distance = 8m, displacement = 8m

b. Distance = 16m, displacements = 0

(i) Graph (c): The speedof the ball goes on decreasing uniformly as it moves upward

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.

a) The first equation of motion is v = u + at 

Where, 

v is the final velocity of the body 

u is the initial velocity of the body 

a is the acceleration 

t is the time taken 

The second equation of motion is s = ut + ½ at² 

Where, 

s is the distance travelled by the body 

t is the time taken 

u is the initial velocity 

a is the acceleration 

The third equation of motion is v² – u² = 2as Where, 

v is the final velocity of the body 

u is the initial velocity of the body 

a is the acceleration 

s is the distance travelled by the body 

b) Initial velocity, u = 0 m/s 

Final velocity, 

v = 72 km/h = 20 m/s 

Time, t = 10s 

i) Acceleration = final velocity – initial velocity/time taken 

a = v-u/t = 2 m/s² 

ii) Average velocity = initial velocity + final velocity/2 

Average velocity = 10 m/s 

iii) Distance travelled = (average velocity)(time) = 100m

Initial velocity = u 

u = 6m/s 

Final velocity = v 

V = 16m/s 

Time = t = 10 s 

• Acceleration = a 

a = \(\frac{(v-u)}{t}\)

=  \(\frac{(16-6)}{10}\)

= 1 m/s² 

• Distance = s

s= \(\frac{(v^2-u^2)}{2a}\)

s = \(\frac{256-36}{2\times1}\)

= 110 m

i) Distance = 100 m; Time = 4 sec

Average speed = \(\frac{100\,m}{4\,sec} = 25 \,m/sec\)

ii) Distance = 50 m ; Time = 2 sec

Average speed = \(\frac{50\,m}{2\,sec}= \,25\,m/sec.\)

The correct answer is d) 1 m.s^-2

Initial velocity of the bus u = 80 kmh^-1

^{80 x \(\frac{5}{18}\) = 22.22 m/s}

^{60 x \(\frac{5}{18}\)} = 16.66 m/s

Final velocity, V = 60 km/h

time taken to decrease the velocity of bus, t = 5 seconds

acceleration, a = 5 seconds

Velocity a = \(\frac{v-u}{t}\)

= \(\frac{16.66-22.22}{5}\)

= -1.112 ms^-2.

Initial speed of the train u = 0

= 40 x \(\frac{5}{18}\) = 11.11 m/s

time = 10 min = 10 x 60 = 600 sec

acceleration a = \(\frac{v-u}{t}\)

= \(\frac{11.11-0}{600}\) = 0.0185 ms^-2

∴ acceleration of a train = 0.0185 ms^-2.

If the speed of the object changing uniformly, magnitude of average velocity of an object equal to its average speed.

When the object is in uniform motion, it covers equal distance at equal intervals of time.

The odometer of an automobile measures the distance covered by an automobile.

Radius (r) = 7 m

Distance covered by object = 2πr =2 × \(\frac{22}7\)×7 =44 m

The object come back to original position displacement of the object is 0.

a) v = u + at is the first equation of motion. 

Where, 

v is the final velocity 

u is the initial velocity 

a is the acceleration 

t is the time taken 

b) Initial velocity, 

u = 0 m/s 

Time, t = 5s 

Distance, s = 100 m 

Acceleration, a = ? 

s = ut + 1/2at²
a = 200/25 = 8 m/s²

False 

Motion along a curved path is not called translator or rectilinear motion. 

Motion along a curved path is called a curvilinear motion. A curvilinear motion is accelerated motion as direction changes.

The correct answer c) in accelerated motion

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

where: v = final velocity 

u = initial velocity = 10 m/s 

a = acceleration = 2.5 m/s² 

t = time = 10s 

v = 10 m/s + 2.5 m/s² x 10s = 10+25 m/s = 35 m/s

 D) \(\cfrac{\sqrt{2l}}t\)

Let “s” be the total distance and “a” be the acceleration. 

Now, we know that initially the body is at rest. 

Now, from the equations of motion, we get, 

\(s=ut+\frac{1}{2}at^2\)…………1 

\(s = \frac{1}{2}a\times4^2\)…………2 

To find the time taken for covering \(\frac{s}{4}\) , we put = \(\frac{s}{4}\) in equation 1 above. 

So, \(\frac{s}{4} =\frac{1}{2}\,a\times\,t^2\) ….…….3 

Now, dividing equation 2 by 3, 

we get, t = 2 s 

Hence, option B is correct.