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Correct option is B) velocity

Correct option is D) zero

Correct option is D) y = 3x

1. a) Body A moves with constant speed since it travels equal distance in equal interval of time. 

b) Body B moves with constant acceleration since the rate of change of distance per second in it is constant. 

2. a) Body C covers the maximum distance in 3^rd second. 

C covers 100 cm whereas A & B cover 20 cm & 24 cm in 3^rd second. 

b) Body A covers the minimum distance in 3^rd second. 

Since A covers only 20 cm which is less than covered by B (24 cm) and C (100 cm) 

3. Body C is moving with non-uniform acceleration

C) at \(\cfrac{1}3^{rd}\) of maximum height is \(\sqrt{\cfrac{2}{3}}\) u

Correct option is C) A and B

Correct option is C) Speed

6 m/s = 6×(3600/1000) km/h = 21.6 km/h.

a) Bus moving on the road 

b) A racing horse motion

Velocity is the name that is given to the speed in a specified direction

The particle is moving in a circular path because at every point the direction is changing along the path and is always tangential to the circular path.

Let the total running time = x hrs

Speed in the first half of its running time i.e., x/2 hrs = 80 km/h

∴ Distance covered in x/2 hrs = 80 x (x/2) =80x/2 = 40x km 

Speed in second half its running time = 40 km/h 

∴ Distance covered in x/2 hrs = 40 x (x/2) = 20x km 

Total distance = 40x + 20x = 60x km

Total time = x/2 + x/2 = x hrs

∴ Average speed  = Total distance /Time  = 60x/x =60 km/hr

We know that, 

Distance = Speed × Time 

So, area under the Speed – Time graph will give the distance covered by the particle. Hence the option C is correct.

We know that  ,\(Distance\propto\,time^2\)     ….1 

Now, \(velocity=\frac{Distance}{Time}\)       ….2 

From equations 1 and 2, we get, Velocity \(\propto\)Time             ….3 

Now, we also know that, \(Acceleration=\frac{velocity}{Time}\)       ….4 

So, from equations 3 and 4, we get, 

\(Acceleration\propto\,time^0\) 

Acceleration = Constant 

Hence, option D is correct

Area under the Velocity – Time graph gives the displacement. 

Now, Displacement = Velocity × time 

Also, area under the graph is calculated by multiplying velocity and time. Thus, displacement is obtained in that process. 

Hence option A is correct.

Speed is not a vector quantity. 

Speed is distance travelled per unit time. It has only magnitude and no direction. Whereas for velocity, displacement and acceleration, all have magnitude as well as direction. 

Hence option C is correct.

If the velocity of the body is always changing with a uniform rate (uniform acceleration), then the average velocity is given by,

\(Average Velocity =\frac{Initial\,Velocity\,+\,Final\,Velocity}{2}\)

\(\bar{v}=\frac{u+v}{2}\)

Hence option D is correct

When velocity of the body changes by unequal amounts in equal intervals of time in a particular direction, the acceleration of the body is said to be nonuniform.

When velocity of the body changes by equal amounts in equal intervals of time in a particular direction, the acceleration of the body is said to be uniform.

When I return to home from school, then my displacement = 0. 

Displacement = Final Position – Initial Position 

Now, initially I was at home and finally also I am at home only. Hence, the displacement = 0

Train A 

Initial Velocity = U_A = 72 kmh^-1 = 20 ms^-1 

Time Taken = t = 50s 

Acceleration = a_A = 0 ms^-2 

Now, from the equations of motion, we know that,

\(s_A =u_At + \frac{1}{2}a_At^2\)

\(s_A =20\times50\)

\(s_A =1000\,m\)

Driver is at starting of Train A 

Train B 

Initial Velocity = U_A = 72 kmh^-1 = 20 ms^-1 Time Taken = t = 50 s 

Acceleration = a_B = 1 ms^-2 

Now, from the equations of motion, we know that,

\(s_B = u_B +\frac{1}{2}a_Bt^2\)

\(s_B =20\times50 +\frac{1}{2}\times1\times50^2\)

\(s_B = 2250\,m\)

Guard is at the end of Train B Now, length of both trains = 400 m + 400 m = 800 m

Now, Original distance between Train A and B is S, which can be obtained as given below: 

S = 2250 m – 1000 m – 800 m 

= 450 m

Given:

Velocity of Car = 18 ms^-1 

To express this velocity in kmh^-1 we need the conversion factor which is calculated as under: 

1 km = 1000 m 

So, 1 m = \(\frac{1}{1000}\) km 

Now, 1 hr = 60 minutes = 60 × 60 s = 3600 s 

So, 1 s = \(\frac{1}{3600}\) hr 

Now, Velocity of car = 18 \(\frac{m}{s}\)

= 18 \(\frac{km}{1000}\) × \(\frac{3600}{hr}\)

= \(\frac{18\times36}{10}\) kmh⁻¹ 

= 64.8 kmh⁻¹

A) uniform motion

C) 50 km / hour

B) c, d, a, b

Correct option is D) Zero

A) Both A and R are correct, R is correct explanation of A

B) varying velocity without having varying velocity

The speed at any instant is called instantaneous speed.

If we apply same force on a car and a motorcycle, then the motorcycle will produce more acceleration. 

We know that, Force = Mass × Acceleration 

Now, force is same on both, but 

Mass of Car > Mass of Motorcycle 

Now, using above equation of force,

F = m × a 

Now, F = Constant 

So, a = \(\frac{f}{m}\)

∝ \(\frac{1}{m}\)

So, acceleration of car < acceleration of motorcycle

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

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.

Correct option is A) 5 min

Uniform motion: When a car moving at a constant velocity of 50Km/hr in a straight line, then the car is covering equal distances in equal intervals of time irrespective of the length of time. Then we can say that the car is in uniform motion. 

Non Uniform motion: When a race car constantly accelerates to win the race, then the car travels unequal distances in equal intervals of time. Hence such an instance can be referred as an example of nonuniform motion. 

Uniform circular motion: When a table fan rotates at a constant speed then it can be described as an uniform circular motion as it is rotating in a circle in which at all instances change of direction is taking place and movement is also circular. 

Non uniform circular motion: Motion of a table fan when its switched off till its stops, can be referred as an example of non-uniform circular motion as the rotation of the fan is not constant and it is changing with time.

Uniformly accelerated Motion: When a body falls freely, then it accelerates uniformly with a magnitude of acceleration due to gravity. Thus velocity of the body increases by equal amounts in equal intervals of time. Hence it can be referred as an example of uniformly accelerated motion. 

Non-uniformly accelerated motion: When a feather falls down in heavy storm, then the velocity of the feather decreases sometimes and sometimes it increases in unequal amounts in equal intervals of time. Thus it can be referred as an example of non-uniformly accelerated motion.

Motion of a rocket, horizontally projected body, kicked football, a cricket ball bowled by a bowler, etc.

Yes, a body can have a constant speed and still be accelerating. 

Acceleration can be either due to change in speed or due to direction of motion or both. 

Consider an example of uniform circular motion. In a uniform circular motion, the body moves with a constant speed but we still say that it is accelerating due to the change in direction. 

This acceleration is called centripetal acceleration and is given as 

a_c = \(\frac{-v^2}{r}\) 

Where, v = constant speed 

And r = radius of circle

Initial velocity, u = 0 m/s 

Time, t = 2s 

Acceleration, a = 10 m/s² 

s = ut + ½ ats = 20m

a) speedometer

b) odometer