Motion

After half a circle., the diameter of the circle will be its displacement.

So, 2r is that answer.

Initial velocity be u

Final Velocity (v)= 0

Acceleration on the body is –g

Hence, from the equation of motion

We have

v² = u² +2as

0 = u² – 2gh

h = u²/2g

So. u²/2g is correct option.

The displacement is always less than or equal to distance.

Its depend on direction of motion.

If motion in same direction than it will be 1 otherwise less than 1.

From the 2^nd equation of motion,

If the object starts from rest i.e. its initial velocity is zero (i.e., u = 0), then

For, s to be proportional to square of time, a must be constant.

Thus, the object moves with uniform acceleration.

The graph shows constant velocity of the object with increasing time. Thus, the object is in uniform motion. So option A is correct.

Acceleration of a body is the rate of change of its velocity with time.

Here the, magnitude of the velocity is constant but its direction is changing. i.e. when the boy is sitting on a merry -go round, the direction of motion is changing, which indicates that the boy is in accelerated motion.

So, option C is right.

The area under the velocity-time graph gives the distance (magnitude of displacement) which has the unit: metre (m)

So, option B is correct.

From the slope of distance-time graph given, we get the speed of the object. Clearly, the slope of B is the lowest and the slope of C is the greatest. So, we can say that the car B is the slowest and thus it will cover the smallest distance in longest interval of time.

For the uniform motion, the slope of distance-time graph is a straight line. Any object is in uniform motion such that equal distance is covered in equal intervals of time.

The slope of any line is given as, Slope = =

And we know, that is Acceleration.

So, Acceleration is the correct answer.

The distance moved and the magnitude of displacement are equal if the car is moving on straight road.

Displacement is the shortest path travelled between any two initial and final points. When a car is moving on a straight road, then the displacement is equal to the distance.

Displacement means the shortest distance that has been travelled by any object in moving process. Displacement will be zero when the initial and the final position of an object are the same.
Distance means the actual path that has been travelled and it is a scalar quantity. For example- if a man walks along a circular path so its displacement is zero but his distance is not zero because it is equal to the circumference.

In uniform motion, So, from 1^st equation of motion are,
v=u + at

If a=0 then

v=u+0 x t

v = u + 0

or v = u

i.e. initial velocity is equal to the final velocity. Or the velocity is uniform.

From 2^nd equation of motion, s=ut+1/2at²

(if a=0)

s=ut+x 0 x t²

s = ut + 0

s = ut

From 3^rd equation of motion,

v² - u² = 2as

(if a=0)

v² - u² = 2 x 0 x s

v² - u²= 0

or v² = u²

Thus, the equations of motion with uniform velocity are:

1) v = u

2) s = ut

3) v² = u²

From the graph, we have:

Initial velocity (u) = 0

Now, Velocity(v) = displacement/time

Therefore, the velocity after 50s =?

v = 100/50 = 2ms^-1

Ans, the velocity after 100s,

Here, the displacement = 0

Then, time = 100s

velocity=displacement/time

V = 0/100 = 0

Velocity – time graph plotted from the above data is:

As the car moves from rest, where u = 0,

Acceleration, a = 5ms^-2

time, t = 8sec

distance(s) =?

From the equation of motion

S₁ = ut + 1/2 at²

S₁ = (0x8) + 1/2 x 5 x (8)²

S₁ = 0 +1/2x5x64

S₁ = 5x32

S₁ =160 m

The velocity after 8 sec,

From equation of motion

v = u + at

v = 0 + 5 x 8

v = 40m/s

So, the distance travelled in 4 sec (12 s – 8 s = 4s)

S₂ = 40 x 4

S₂ = 160 m

Therefore, total distance travelled by the car = S₁ + S₂

= 160 + 160

= 320 m

Let the distance (AB) = x

Total distance = x + x = 2x

Speed = distance/time

t₁ = x/30

t₂ (return trip) = x/20

Total time taken = t₁ +t₂

Total time taken = x/30 + x/20

= (2x +3x)/60

= 5x/60

= x/12

Average speed = total distance/total time taken

= 2x /

= (2x) x

= 24 km/hr

(i) From the graph, the velocity is constant with repsect to time i.e. the velocity is not changing and thus, rate of change of velocity is zero i.e. acceleration is zero.
(ii) The velocity is same as throughout as the it is constant with time. According to graph, Velocity = 20ms^-1

(iii) Distance covered covered by the cyclist in 15 seconds,

S = u × t =20 m/sec x 15 sec = 300 m

let h₁=150m and h₂=100m

Then, the difference in height h=(150-100)m=50 m

Distance travelled by first body in 2s is given as,

Distance=ut + at²

Now, if the Initial speed(u) = 0(when the object was at rest)

Ans, the time, t =2s

So, by substituting the values, we get,

S = 0 +g× (2)²

S = 0 +g× 4

S = 0 +× 9.8 × 4

S = 19.6 m

So, After 2s, the height of the first object, h₁=150-19.6

So, After 2s height of the second object, h₂=100-19.6

Thus, after 2s difference of height= h₁ - h₂ (150-19.6) - (100-19.6) =50 m

So, after 2s difference height will be 50 m = initial difference in height

Thus, the difference in height of the object will remain same because the height does not vary with time.

Let acceleration (a) of the body be ‘a’.

So, to find the acceleration of the body

..............(i)

Where u = 0, t = 2s and s = 20m

So, by putting the values in equation (i), we get,

20 = 0 +x a x(2)²

20 = x 4a

a = 40/4

Therefore the velocity after 2 seconds is :-

v= 0 + 10×2= 20m/s

We know that the distance travelled in next 4 seconds is 160 m and the body is starting at 20m/s

∴ u=20m/s
t= 4 sec
s= 160m

Puttting the above value in

We get :-

Solving for a^' we get a^'= 10 m/s²

Since the acceleration is same , we use

V= u + at

V = 70 m/s

Average velocity =

Or V=1ms^-1

For next 4s,

V = 0/4

V = 0 ms^-1

As X(metre) remains the same from 4 to 8 seconds, velocity = 0

For last 6s,

V=

V = -6/6

V = -1ms^-1

given

Initial velocity u = 5x10⁴ m s^-1

And acceleration a = 10⁴ m s^-2

(i) According to the given question, the final velocity v = 2u

We need to find the time, t =?

V = u+ at

2u = u +(10⁴ ms^-2 ) × t

u = 10⁴ ms^-1 × t

t = u / 10⁴ m s^-1

t = (5x10⁴) / 10⁴

=5s

(ii) t=5s, a=10⁴ ms^-2, u=5x10⁴ms^-1 s=?

s=ut + at²

s = (5 x10⁴) x5 +1/2(10⁴) x (5)²

s = 25 x10⁴ + 25/2 x (10⁴)

s = 25 x 10⁴ +12.5 x (10⁴)

s = 37.5x10⁴ m

from the equation of motion

s= ut + at² Distance travelled in 5s,

s = u x 5+ x a x (5)²

s =5u + 25/2 a …….(i)

Similarly, distance travelled in 4s,

s’ =4u +16/2 a

s’ =4u+8a…….(ii)

Distance travelled in the interval between 4^th and 5^th second=(s-s’)

= (5u +25/2 – 4u +8)

= (u+ a) m

We know for upward motion,

v²= u² - 2gh

so,

h=v² -u²/2g

Highest point v=0

so,

H=u²/2g

Height for first ball, h₁= u₁²/2g

Height for second ball, h₂= u₂²/2g

Then, h₁/h₂= (u₁²/2g) /(u₂²/2g)

= u₁²/u₂²

= h₁:h₂

=u₁²: u₂² Proved