Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

\(\vec{v}=2 \hat{l}+t^{2} \hat{j}-9 \vec{k}\)
\(\vec{a}=\frac{d \vec{v}}{d t}=2 t \hat{j}\)
\(\text { When } t=0.5 \mathrm{~s}\)
\(a=1 \mathrm{~ms}^{-2}\)

Velocity and acceleration are in the same direction: So speed increases.

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

When the projectile hits the ground after initially thrown horizontally from the top of tower of height h, the time of flight is
\(t=\sqrt\frac{2h}{g}\)
The horizontal component velocity of the projectile remains the same i.e v_x = u.
The vertical component velocity of the projectile at time T is

\({ v }_{ y }=gT=g\sqrt { \frac { 2h }{ g } } =\sqrt { 2gh } \)
The speed of the particle when it reaches the ground is
\(v=\sqrt { { u }^{ 2 }+2gh } \)

(i) t = \(\theta\) to t = 3:
velocity at t = 0, u = 0
velocity at t = 3 see, v = 30 mls
So, from velocity v = u + at
Acceleration \(a={{u}\over{t}}={{30}\over{3}}=10{ms}^{-2}\)
So distance covered between 0 to 3 see
\(s=ut+{{1}\over{2}}{at}^{2}\)
initial velocity u = 0 and t = 3
\(s=ut+{{1}\over{2}}{at}^{2}\)
\(=0\times3+{{1}\over{2}}\times10{(3)}^{2}\)
\(=0+{{1}\over{2}}\times10\times9\)
\(={{1}\over{2}}\times90\)
s = 45 m
From 3 to 6 see, velocity is same 30 m/s.
So distance travelled, = 30\(\times\)3 = 60. m
Total distance covered between 0 to 6 sec.
= 30 + 60
= 90m
(ii) At t = 3 sec, u = 30 ms^-1, at t = 6 see,
v = 0 m/s .
So from, v = u + at
0= 30 + a\(\times\) 3
a =-10m/s²
So, distance covered from t = 3 to t = 6 see is,
v² = u² +2as
(0)² = (30)² - 2\(\times\)10\(\times\) s
\(s={{900}\over{20}}=45\ m\)
So total distance covered = 90 + 45
= 135 m

Initial velocity 'u' =0
Acceleration 'a' = 0.2 ms^-2
Time (t) = 3 minutes = 180 seconds.
Speed of a bus, v = u + at
= 0 + 0.2\(\times\) 180
= 36 m/s.
Displacement s = ut+\(\frac { 1 }{ 2 } { at }^{ 2 }\)
\(=0\times 180+\frac { 1 }{ 2 } \times 0.2\times ({ 180) }^{ 2 }\)
\(=0+\frac { 1 }{ 2 } \times 0.2\times 32400\)
= 3240 m
=3.24 km

Total distance to be covered = 1 km + 100 m = 1100 m (including both bridge and time)
Then, Speed=60 kmh^-1\(=60\times\frac{5}{18}ms^{-1}=\frac{50}{3}\ ms^{-1}\)
Then, time taken to cover this distance \(=\frac{1100}{\frac{150}{9}}s=66s\)

The echo will be heard when the sound reaches back at the place of firing. So, the total distance travelled by sound is 2\(\times\)1.2 km = 2.4 km = 2400 m.
\(speed=\frac{2400m}{8s}=300\ ms^{-1}\)

We know that when the object falls towards the Earth, it experiences acceleration due to gravity g = 9.8 m s^-2 downward. We can choose the coordinate system as shown in the figure. The acceleration is along the negative y direction.
\(\vec { a } =g(-\hat { j } )-g\hat { j } \)

Average velocity: The average velocity is defined as ratio of the displacement vector to the corresponding time interval.
\(\vec{v_{avg}}=\frac{\Delta\vec{r}}{\Delta t}\)
It is a vector quantity. The direction of average velocity is in the direction of the displacement vector (AY).
Average speed: The average speed is defined as the ratio of total path length travelled by the particle in a time interval.
Average speed = Total path length / total time
It is a scalar quantity. Doesn't possess direction.

Area = \(\overset { b }{ \underset { a }{ \int } } f(x)dx=\lim _{ \Delta x\rightarrow 0 }{ \overset { N }{ \underset { i=1 }{ \Sigma } } } f\left( { x }_{ i } \right) \triangle x\)

Any physical quantity is represented by a "function" in mathematics. Take the example of temperature T. We know that the temperature of the surroundings is changing throughout the day. It increases till noon and decreases in the evening. At any time "t" the temperature T has a unique value. Mathematically this variation can be represented by the notation 'T (t)' and it should be called "temperature as a function of time". It implies that if the value of 't' is given, then the function "T (t)" will give the value of the temperature at that time 't'.
Eg: Consider a function f(x) = x² Sometimes it is also represented as y = x². Here y is called the dependent .variable and x is called independent variable. It means as x changes, y also changes. Once a physical quantity is represented by a function, one can study the variation of the function over time or over the independent variable on which the quantity depends.

Given:
Speed of a particle v = 10t²m/s
Formula:
\(s=\int { v } dt\)
\(=\int _{ 2 }^{ 5 }{ 10{ t }^{ 2 }dt } =10\int _{ 2 }^{ 5 }{ { t }^{ 2 }dt } \)
\(=\frac { 10 }{ 3 } \left( { t }^{ 3 } \right) _{ 2 }^{ 5 }=\frac { 10 }{ 3 } \left( { 5 }^{ 3 }-{ 2 }^{ 3 } \right) =390m\)

Given:
t=3s
Distance s = Area of \(\triangle OAB\)

\(=\frac{1}{2}\times OA\times BA\)
\(=\frac{1}{2}\times 3\times 6=9m\)
If the speed varies with the time: Then \(v=\frac{ds}{dt}\Rightarrow ds=v\ dt\)
\(\Rightarrow \int { ds } =\int { v }\ dt\)
\(or\ s=\int { v }\ dt\)

Given: Total distance = s
Velocity of the girl for the first half distance = v₁
Velocity of the girl for the second half distance = v₂
Formula:
\(Time\ taken=\frac{Distance}{Velocity}\)
Time taken for the first half distance \(=\frac{\frac{s}{2}}{v_{1}}=\frac{s}{2v_1}\)
Time taken for the second half distance \(=\frac{\frac{s}{2}}{v_{2}}=\frac{s}{2v_2}\)
Total time taken \(=\frac{s}{2v_1}+\frac{s}{2v_2}\)
Average speed,\(v_{av}=\frac{s}{\frac{s}{2v_1}+\frac{s}{2v_2}}=\frac{2v_1v_2}{v_1+v_2}\)

The relation between linear and angular velocity is given by
v=rω
Differentiating the equation with respect to time, we get (since r is constant)
\(\frac{dv}{dt}=\frac{rd\omega}{dt}=r\alpha\)
Here, \(\frac{dv}{dt}\) is the tangential acceleration and is denoted as \({ a }_{ t }.\frac { d\omega }{ dt } \) is the angular acceleration \(\alpha\). Then
\(at=r\alpha\)

The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight.
This time of flight is the time taken by the projectile to go from point O to B via point A in the figure (Initial velocity resolved into components)

We know that \({ s }_{ y }={ u }_{ y }t+\frac { 1 }{ 2 } { a }_{ y }{ t }^{ 2 }\)
Here, S_y=y = 0 (net displacement in y-direction is zero), u_y = u sinθ, a_y = -g, t = T_f
\(0=usin\theta { T }_{ f }-\frac { 1 }{ 2 } g{ T }_{ f }^{ 2 }\)
\({ T }_{ f }=2u\frac { sin\theta }{ g } \)