Option 4 : *(v-u) / t*

The correct answer is __ (v-u) / t__.

__ CONCEPT__:

**Acceleration**: The rate of change in velocity is called acceleration. It is denoted by a.- The SI unit of acceleration is m/s
^{2}.

- The SI unit of acceleration is m/s
- Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
- V = u + at
- Where, V = final velocity, u = initial velocity, a = acceleration of body under motion, and t = time taken by the body under motion.

__ EXPLANATION__:

From the **above equation of motion**:

V = u + a t

So a = (V - u)/t

- The
**formula for ﬁnding acceleration is (v-u) / t.**So option 4 is correct.

__Additional Information__

- The product of velocity and time
*(v*x*t)*represents the**displacement of a body**which represents the shortest distance between two points. - The reciprocal of time
*(1 / t)*represents the**frequency of oscillations**which gives the value of the total number of oscillations in the given time. - The work per unit time
*(W / t)*represents**power**which is the measurement of the rate of doing work.

A ball is dropped from a height of 80 m. The distance travelled by it in the fourth second will be ________.

(take g = 10 m/sOption 4 : 35 m

__ CONCEPT__:

The **distance travelled by a body in n ^{th} second** is given by:

Sn = u + a/2 (2n - 1)

Where u is the initial velocity of the body, a is acceleration, and n is nth second.

__ CALCULATION__:

Given that:

The ball is dropped from some height, so initial velocity (u) = 0 m/s

Acceleration is due to gravity (a) = 10 m/s^{2}

and n = 4

**For nth second**, S_{n }= u + a/2 (2n - 1)

Distance travelled in 4th second (S_{4 }) = 0 + 10/2 (2 × 4 - 1) = 35 m

- Therefore, the
**distance travelled by the ball in the fourth second will be 35 m.**So option 4 is correct**.**

Option 3 : decreases

The correct answer is option 3, Decreases.

__CONCEPT__:

- Velocity: The rate of change of displacement of a body is called the velocity of that body.
- Velocity is a vector quantity that has both magnitudes as well as direction.

- Acceleration: The rate of change of velocity is called as the acceleration of the body.
- Acceleration is also a vector quantity.

- Retardation/Negative acceleration: When the acceleration of the body is in the opposite direction of the velocity is called deceleration.
- It is a negative acceleration of the body.

__EXPLANATION__:

- The direction of acceleration is determined by the general principle that if an object is slowing down, the acceleration is in the opposite direction of the motion.
- Hence, with
**negative acceleration,**the**velocity**of a body**decreases**. - The negative sign of acceleration shows that the velocity of the body is
**decreasing per unit of time**.

Option 3 : 150 metres

__CONCEPT__:

- Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
- These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance travelled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

__CALCULATION__:

Given that:

Mass of body ( m ) = 10 kg

Acceleration (a) = 3 m/sec^{2}

Time (t) = 10 seconds

Since, \({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

As the body starts from rest u = 0 i.e. Initial velocity (u) = 0

S = 0 + (1/2) × 3 × 10^{2 }

S = 0.5 × 3 × 100

S = 150 metres.

⇒** Distance is 150 meters.**

Option 2 : 3 s

__CONCEPT__:

- Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
- These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

__CALCULATION__:

Given that:

Acceleration (a) = 20 m/s^{2}

Distance traveled (S) = 90 m

Initial Velocity (u) = 0

To find the time taken (t) to cover 90 m.

Equation of motion states that:

\(S=ut+\frac{1}{2}at^2 \)

\(90=0\times t+\frac{1}{2}\times 20 \times t^2\)

\(t^2=9\)

**Time taken (t) = 3 sec**

Hence option 2 is correct.

Option 1 : 2 ms^{-2}

__ CONCEPT__:

- These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

__ CALCULATION__:

It is given that,

⇒ Initial Velocity (u) = 36 km/hr = (36 × 1000 m)/3600 sec = 10 m/s

⇒ Final Velocity (V) = 72 km/hr = (72 × 1000 m)/3600 sec = 20 m/s

⇒ Time (t) = 5 Sec

**Use V = u + a t**

20 = 10 + a × 5

⇒ Acceleration (a) = 10/5 =** ****2 **ms^{-2}

So option 1 is correct.

Option 4 : 75

__CONCEPT__:

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

__CALCULATION__:

Given that:

Initial velocity (u) = 10 m/s

Acceleration (a) = 2 m/s^{2}

Time (t) = 5 sec

S = ut + 1/2 at^{2}

**Displacement (S) = 10 × 5 + (1/2) × 2 × 5 ^{2} = 75 m**

Hence option 4 is correct.

Option 4 : Acceleration

The correct option is 4.

__CONCEPT__:

- Velocity (v): The rate of change of position i.e. rate of displacement with time is called velocity.

\(v = \frac{{{s_2} - {s_1}}}{{{t_2} - {t_1}}} = \frac{{{\rm{\Delta }}s}}{{{\rm{\Delta }}t}}\)

Where s2 = displacement of the object at t2 and s1 = displacement of the object at t1

- It is a vector quantity.

__EXPLANATION__:

- The slope of any graph is the ratio of the vertical change between two points to the horizontal change between the same points.
- In a velocity-time graph, the velocity (v) is present on the vertical axis while time (t) on the horizontal axis, so the slope of the graph is given by

\({\rm{Slope}} = \frac{{{\rm{\Delta }}v}}{{{\rm{\Delta }}t}}\)

- Since the rate of change of velocity is termed as acceleration, the slope of a velocity-time graph gives the acceleration.

Option 4 : Zero

The correct answer is __ option 4__ i.e

**CONCEPT:**

**Velocity (v)**: The rate of change of displacement is called velocity.**Uniform velocity:**The motion in which a body is moving with a constant velocity (no acceleration) is called uniform velocity motion.**Acceleration (a):**The 'Rate of velocity change is called acceleration.- When velocity either increases or decreases over time or changes its direction, acceleration is produced which is positive if it increases and negative if it decreases, and this also depends on the direction of reference that we have taken positively.

- Therefore, as in this case velocity is constant,
**its direction and magnitude are constant.**There is no acceleration there.

a = (v_{2} - v_{1})/(t_{2} - t_{1})

If the change in time is infinitesimally small then we have the differential operator representation:

Acceleration (**a) = dv/dt **

⇒ dv = change in velocity

⇒ dt = change in time

__ EXPLANATION__:

Now at constant quantity: zero acceleration

- When
**an object is moving with a uniform velocity, its acceleration is Zero.**So option 4 is correct.

Option 1 : The liquid surface falls down on the direction of motion and rises up on the back side of the tank

__CONCEPT__:

- Acceleration: The rate of change of velocity is called the acceleration of the body. i.e.,

\(a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\)

Where v2 = velocity of the object at t2 and v1 = velocity of the object at t1

- It is a vector quantity.
- Its direction is the same as that of change in velocity (Not of the velocity).

__EXPLANATION__:

- As we know that if the liquid is accelerated horizontally, then the two points in the same horizontal line can’t have equal pressure.
- Therefore, the free surface of the liquid is inclined to the horizontal with larger depth at the rear end as the tanker is moving in the forward direction with uniform acceleration. Hence option 1 is correct.

Option 1 : 6 m/s

__CONCEPT__:

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

__CALCULATION__:

Given that:

a = 2 m / s^{2}

t = 3s

v = ?

**For initial speed, u = 0**

v = u + at = 0 + 2 × 3

**v = 6 m**

Hence option 1 is correct.

Option 1 : 62.5 ms^{-1}

__CONCEPT__:

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

__EXPLANATION__:

Here acceleration is acceleration due to gravity ( a = g = 10 m/s2).

Time of ascent (t) = Time of descent (t)= 12.5/2 = 6.25 s

Final velocity (v) = 0 (**at maximum height**) , Initial velocity (u) = ?

Using v = u + at,

we get

⇒ 0 = u + (-10 x 6.25)

⇒ -u = - 10 x 6.25

⇒ u = 62.5 ms^{-1}

**The Velocity with which it was thrown up is****62.5 ms-1**. So option 1 is correct.

Option 2 : Uniform

The correct option is: 2

__CONCEPT__:

- Velocity (v): The rate of change of displacement of a body is called the velocity of that body.
- Velocity is a vector quantity that has both magnitudes as well as direction.

- Acceleration (a): The rate of change of velocity is called the acceleration of the body.
- Acceleration is also a vector quantity.
- The slope of any velocity-time graph gives an acceleration of the body

a = dv/dt

Velocity (v) = dx/dt

Where x is displacement and t is time

- Uniform acceleration: When the acceleration is constant then it is called uniform accelerated motion.
- Non-uniform acceleration: When the acceleration is not constant then the motion is non-uniform accelerated motion.

__ CALCULATION__:

Given that:

v2 = a + bx

Differential both sides with respect to x,

2v (dv/dx) = 0 + b × (dx/dx)

\(v\frac{dv}{dx} = \frac{b}{2}\)

Since a = dv/dt, and Velocity (v) = dx/dt

Now \(a = \frac{dv}{dt} =\frac{dv}{dx} × \frac{dx}{dt} = v\frac{dv}{dx}\)

Hence **acceleration (a) = v(dv/dx) = b/2**

**Since b is constant so acceleration will be constant and hence uniform acceleration.**

Option 1 : 1 sec

__CONCEPT____:__

- Equation of Kinematics: These are the various relations between u, v, a, t and s for the particle moving with uniform acceleration where the notations are used as:
- Equations of motion can be written as

V = U + at

\(s =ut+\frac{1}{2}{at^{2}}\)

V2 =U2+ 2as

Where, U = Initial velocity, V = Final velocity, g= Acceleration due to gravity, t = time, and h= height/Distance covered

Where u = Initial velocity of the particle at time t = 0 sec

v = Final velocity at time t sec

a = Acceleration of the particle

s = Distance travelled in time t sec

__CALCULATION__:

Given - Initial velocity (u) = 490 cm/s = 4.9 m/s

When a particle is thrown **vertically upward**, then at maximum height final velocity is zero i.e., v = 0 m/s

- The time taken to reach the maximum is

⇒ v = u + at_{1}

\(⇒ t _1= \frac{u}{g}=\frac{4.9}{9.8}=0.5 sec\) ------- (1)

- The
**maximum height reached by a particle**when it is thrown vertically is

⇒ v2 = u2+ 2as

⇒ -v2= - 2gs

\(⇒ s = \frac{u^2}{2g}=\frac{4.9\times 4.9}{2\times 9.8}=1.225 \,m\)

- The time taken by the particle to reach its initial position is

\(⇒ t^2_2 =\frac{2s}{g}\)

\(⇒ t^2_2 =\frac{2\times 1.225}{9.8}=0.25 \, sec\)

⇒ t_{2} = 0.5 sec ------ (2)

- The total time taken to return to its initial position is

⇒ T = t_{1} + t_{2} = 0.5 + 0.5 = 1 sec

Option 1 : 15 m

The correct answer is __ 15 m__.

**Concept: **

- There are three equations of motion:
- V = u + at
- V2 = u2 + 2 a S
- \({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)
- Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion

__ CALCULATION__:

⇒Given that: Initial velocity (u) = 20 m/s and time (t) = 3 sec

**We Know that,**

**⇒ v = u + at**

⇒ v = 20 - 10t

⇒ **t = 2 s**

The maximum height of Ball = u^{2}/2g

= (20^{2})/2 * 10

= 20 m

After 2 s ball reaches heighest position, u =0

⇒ **S = ut + 0.5 at ^{2}**

⇒ S = 0 - 0.5 * 10 * 1^{2}

⇒ S = 0 + 5

⇒ S = -5 m

so, displacement in 3 sec = 20 m - 5 m = **15 m.**

Option 1 : 1 ms^{-2}

__CONCEPT__:

- These equations are only valid when the
**acceleration of the body is constant**and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

__ CALCULATION__:

It is **given that**,

⇒ Initial Velocity (u) = 18 km/hr

⇒ Final Velocity (v) = 36 km/hr

⇒ Time (t) = 5 Sec

**Use V = u + at**

⇒ Acceleration (a) = Change in Velocity (V-u)/Time (t)

⇒ Change in Velocity = (v - u) = 36 - 18 = 18 km/h = 18 × 5/18 = 5 m/s

⇒ Time (t) = 5 Seconds

⇒ Acceleration (a) = 5/5=** 1 ms ^{-2}**

- When we need to
**convert km/h into m/s then multiply the speed with 5/18.** - When we need to convert m/s into km/h then multiply the speed with 18/5.

Option 2 : V_{av} = (u + v)/ 2

__ CONCEPT__:

- Average velocity: The ratio of net displacement to total time taken is called average velocity.

\(Average\;velocity\;\left( {V_{av}} \right) = \frac{{{\rm{Net\;displacement\;}}\left( {{\rm{S}}} \right)}}{{time\;taken\;\left( t \right)}}\)

The average velocity is also given by:

V_{av} = (u + v) / 2

Where u is initial velocity and v is the final velocity of the object.

__EXPLANATION__:

If final Velocity v and Initial velocity u are known, then the formula for the **average velocity **is given by:

**V _{av} = (u + v)/2.**

So option 2 is correct.

Option 4 : 180 m

** CONCEPT**:

**Equation of motion:**Equations of motion relate the displacement of an object with its velocity, acceleration, and time.- The
**motion of an object**can follow many different paths. Here the**motion is in a straight line**(one dimension).

- The
- When the body is displaced with
**negative acceleration.**

** v = u + at**

\(s= ut +\frac{1}{2}at^{2}\)

** CALCULATION**:

Given:

v = 0 (since the velocity will become zero at the highest point)

u = 60m/s

a = -10m/s (acceleration due to gravity and "-" sign for the opposite direction.

using the the formula:

**v = u + at**

0 = 60 -10t

t = 6 sec.

- Now we know after t = 6 sec the ball will be at the highest point.

\(s= ut +\frac{1}{2}at^{2}\)

\(s = 60 \times 6 + \frac{1}{2}\times -10\times 6 \times 6\)

\(s = 180m\)

- The
**maximum height attained by the particle will be 180m**. - Hence option 4 is the answer.

Option 1 : Velocity

The correct answer is **Velocity.**

__ CONCEPT__:

**Velocity**: The rate of change of displacement of a body is called the velocity of that body.- Velocity is a vector quantity that has both magnitudes as well as direction.

**Acceleration**: The rate of change of velocity is called the acceleration of the body.- Acceleration is also a vector quantity.

**Retardation/Negative acceleration**: When the acceleration of the body is in the opposite direction of the velocity is called deceleration.- It is a negative acceleration of the body.

__ EXPLANATION__:

- The direction of acceleration is determined by the general principle that if an object is slowing down, the acceleration is in the opposite direction of the motion.
- Therefore, from this principle, it becomes clear that when an object is slowing down, the
**acceleration is in the opposite direction of velocity. This is a negative acceleration**. So option 1 is correct.

Option 3 : 4 : 3

The correct answer is option 3) i.e. 4 : 3

__CONCEPT:__

Kinematic equations of motion:

- These equations define the relationship between initial velocity u, final velocity v, time t, and displacement s of an object with respect to its motion in uniform acceleration a.
- Following are the three kinematic equations for uniformly accelerated motion:

⇒ v = u + at

⇒ s = ut + 0.5at2

⇒ v2 - u2 = 2as

__CALCULATION:__

Given that:

Fall distance of the objects, h_{1} = 64 m and h_{2} = 36 m

Let the time taken by the two objects to reach the ground be t_{1} and t_{2}.

Since the object is falling freely, acceleration = g

The initial velocities of both objects will be zero ⇒ u = 0 m/s

Using,

⇒ s = ut + 0.5at^{2}

- For an object of mass 2 kg, the time taken by them to reach the ground is

⇒ h_{1} = 0 + 0.5gt_{1}^{2} ------ (1)

- For an object of mass 7 kg, the time taken by them to reach the ground is

⇒ h2 = 0 + 0.5gt22 ------- (2)

On dividing equations 1 and 2, we get

\(⇒ \frac{h_1}{h_2} = \frac{t_1^2}{t_2^2}\)

\(\Rightarrow\frac{t_1}{t_2}= \sqrt{\frac{h_1}{h_2} } = \sqrt{\frac{64}{36} } =\frac{8}{6} = \frac{4}{3} \)