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a) Acceleration 

b) Retardation 

c) No, because when a body takes a round trip from its final position to the initial position, then the distance travelled will be a non-zero whereas the displacement is zero

The speed of a moving body along a circular path is given as: 

v = (2πr)/t 

Where, 

v is the velocity 

r is the radius of the circular path 

t is the time taken to complete one round

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.

For bus X, Speed = distance/time Speed = 360/5 = 72 km/h For bus Y, Speed = 476/7 = 68 km/h Therefore, the speed of bus X is more than the bus Y

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

b) u = 0 m/s v = 21 m/s t = 1 min = 60s a = v-u/t = 0.35 m/s²

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

C) both A and B

Relative: That exists or that has a particular quality only when compared with something else. 

Ex: The position of the sun is relative to the earth.

Distance:

• Distance is the length of the path traversed by an object in a given time Interval.
• Distance is a scalar quantity.

Displacement

• Displacement is the shortest distance covered aby the object in a specified direction. Its unit in the SI system is ‘meter’.
• Displacement is a vector quantity.

Correct option is A) Zero

To know how slow or fast the body is moving, speed is used.

A) average velocity

D) All the above

1. Acceleration gives an idea how quickly velocity of a body changing. 

2. It is equal to the rate of change in velocity. 

3. While travelling in a bus or car, when the driver presses the accelerator, the passengers sitting in the bus experience acceleration. Their bodies press against the seats due to acceleration.

A body is said to have a uniform velocity when it covers equal distances in equal intervals of time in a particular distance.

The resultant displacement of a body is said to be zero when the body comes back to its starting point and covers a certain non-zero distance.

D) A → B → C → A

Correct option is C) A or B

Along with speed, the direction of the body should be known.

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

A) uniform motion

When we apply brakes to the car, then the direction of the acceleration is opposite to that of the motion of car. 

Let the car is moving in positive x direction. Now, we apply brake to stop the car. Now for stopping the car force must act on it (Newton’s First Law) in opposite direction. Hence acceleration is in negative x direction. We may also call it as retardation.

C) Mud will stick to the wheels and speed decreases.

A) instantaneous speed

C) retardation

Correct option is A) 25√2

Yes.

1. The indicator in speedometer changes its position even for a small change in speed. 

2. As it indicates a constant reading, the car moves equal distances at equal intervals of time. 

3. Hence the motion is uniform.

Maximum height reached by the body h=\(\frac{u^2}{2g}\)=\(\frac{4\times4}{2\times10}\)=0.8 m.

In physics, motion, can be classified as below. 

1. Linear motion: Motion along a straight line. 

2. Circular motion: Motion along a circular path. 

3. Oscillatory motion: Repetitive to and fro motion of an object at regular interval of time. 

4. Random motion: Motion of the object which does not fall in any of the above categories. 

5. Uniform and Non-uniform motion:

• Uniform motion: Consider a car which covers 60 km in the first hour, 60 km in second horn-, and another 60 km in the third hour and so on. The car covers equal distance at equal interval of time. We can say that the motion of the car is uniform. An object is said to be in uniform motion if it covers equal distances in equal intervals of time howsoever big or small these time intervals may be
• Non-uniform motion: Now, consider a bus starting from one stop. It proceeds slowly when it passes through a crowded area of the road. Suppose, it manages to travel merely 100 m in 5 minutes due to heavy traffic and is able to travel about 2 km in 5 minutes when the road is clear. Hence, the motion of the bus is non-uniform i.e. it travels unequal distances in equal intervals of time.

C) motion in opposite direction

Given: 

height = 20 m 

acceleration = 10 ms 

Formula: 

For free falling body,

1. v = gt 

2. s = 1/2 gt²

time taken to strike the ground 

s = 1/2 gt² 

20m = 1/2 × 10 ms^-2 × t²

t² = \(\frac{40 m}{10 ms^{-2}}\)= 4s² 

∴ t = 2s 

3. Velocity of the ball when it strikes ground v = gt

v = 10ms^-2 × 2s 

v = 20ms^-1

Velocity changes. 

As the ant is moving on the surface of a ball, it has to go in circular motion.

The bus B covered more distance.

b. Speed = \(\frac{distance}{time}\)

Speed of A = \(\frac{75}{5}\) = 15 m/s

Speed of B = \(\frac{169}{13}\) = 13 m/s

So the bus A has higher speed

Velocity = 300 km/h =  \(300\,\times\,\frac{5}{18}\,=\frac{500}{6}=83.33\,m/sec\)   

As the velocity is constant, the acceleration is also constant.

∴ Acceleration = 83.33 m/sec² .

Distance ball covers, S = 20 m
acceleration, a = 10Kmh^-1 = 10 m/s
initial velocity, u = 0 (at rest)
When ball strikes, its final velocity,
According to 3rd law of motion
v² = u² + 2as
v² = 0 + 2(10)(20)
v = 20 m/s.
According to 1 st Law of motion v = u + at
Time taken by ball to strike the ground, t,
20 = 0 + 10 (t)
∴ t = 2s
∴ 2s is required to ball for striking the ground and velocity is 20 m/s

Uniform acceleration: The motion of an object along a straight line with constant acceleration. 

(OR)

Equal chances of velocity occur In equal Intervals of time.

\(Acceleration=\frac{Final\,velocity-Initial\,velocity}{time}\)

Acceleration = \(\frac{40-0}{8}\) = 5 m/s²

Airplane :  Change in velocity = 5 km/h = \(5\,\times\,\frac{5}{18}= \frac{25}{18}m/sec.\)

Time =10 sec.

Acceleration = \(\frac{{\frac{25}{18}}}{10}=\frac{25}{18}\,\times\,\frac{1}{10}=\frac{5}{36}=0.138\,m/sec.\)

Skateboard : Change in velocity = 5 km/h =\(\frac{25}{18}\,m/sec\)

Time = 1 sec.

Acceleration = \(\frac{{\frac{25}{18}}}{1}=\frac{25}{18}=1.38\,m/sec^2\)

 \(\therefore\) Skateboard has greater acceleration.

Correct option is C) m/s²

a) Shade lightly in the graph about distance travelled by car in 4s.
b) Shade with red colour about 6 cm to 10 cm. It represents uniform motion of the car.

Velocity = \(\frac{dispalcement}{time}\)

Displacement = velocity × time = 15 × 30 = 450m

Average speed = \(\frac{Total\,distance\,travelled}{time\,taken\,to\,travel\,the\,distance}\)

Time taken to travel the distance 400m = \(\frac{400}{8}\) = 50s

Time taken to travel the distance 1200m = \(\frac{1200}{10}\) = 120s

Time taken to travel the distance 360m = \(\frac{360}{12}\) = 30s

Total distance travelled = 400 + 1200 + 360 = 1960m

Total time to travel = 50 + 120 + 30 = 200s

Average speed = \(\frac{1960}{200}\) = 9.8 m/s

Acceleration = \(\frac{0-30}{5}\) = 6m/s²

∴ Retardation = 6m/s²

(a) Possible

When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s².

(b) Possible

When a car is moving in a circular track, its acceleration is perpendicular to its direction.