Initial velocity  ( u ) = 0

Acceleration (${\mathrm{\alpha }}_1$ ) = 2 m/s² 

time during acceleration (t₁) = 10 sec

Time during constant velocity (t₂ ) = 30 sec

and retardation ( ${\mathrm{\alpha }}_2$ ) = $-$ 4 m/s² 

( negative sign due to retardation )

Distance covered by the particle during acceleration

    s₁ = ut₁ + $\frac{1}{2}$ ${\mathrm{\alpha }}_1$${\mathrm{t}}_1^{ 2}$

         = ( 0 × 10 ) + $\frac{1}{2}$ × 2 × (10)²

   s₁ = 100 m                 ......(i)

And velocity of the particle a the end of acceleration 

  v = u  + $\frac{1}{2} {\mathrm{\alpha }}_1{\mathrm{t}}_1^{ 2}$   

      = 0 + ( 2 × 10 ) 

   v = 20 m/s

Therefore distance covered by the particle during constant velocity

  s₂ = v × t₂ 

      = 20 × 30⇒

  s₂ = 600 m                    .....(ii)

Relation for the distance covered by the particle during retardation ( s₃ ) is

   v² = u² + 2${\mathrm{\alpha }}_2$s₃

⇒   ( 0 )² = ( 20 )² + 2 × ( $-$ 4 ) × s₃

⇒    400 = 8 s₃

⇒   s₃ = $\frac{400}{8}$

⇒   s₃ = 50 m

Therefore total distance covered by the particle 

  s = s₁  + s₂ + s₃

    = 100 +600 + 50

 s = 750 m

For a body moving with constant acceleration $\mathrm{\alpha }$

      v = u + $\mathrm{\alpha }$t

Since the body starts from rest

      u = 0

∴     v = $\mathrm{\alpha }$t

which is a straight line passing through the origin. Hence the graph is (b)

Given:- radius of earth=6400km=6400$\times$10³, g=10m/s²

g^'=0, m=0(weight is Zero at equator)

g=R$\omega$²  $\Rightarrow$$\omega$²=$\frac{g}{R}$

Angular speed$\omega$=$\sqrt{\frac{\mathrm{g}}{\mathrm{R}}}$=$\sqrt{\frac{10}{6400\times {10}^3}}$=$\frac{1}{800}\mathrm{rad}/\sec$

Kilometres per hour is a unit of measurement, which measures speed or velocity. The unit  symbol is km/h  or km h^-1.

By definition, an object travelling at a speed of 1 km/h in a hour moves 1 kilometre.

Angular acceleration= 2 rad/s²

Time interval t= 10sec

Angular displacement θ = ?

$$\begin{gathered} N\mathrm{ow} \mathrm{by} \mathrm{\theta } =\mathrm{\omega t}+\frac{1}{2}{\mathrm{\alpha t}}^2 ....\left(\mathrm{i}\right) \\ \left[\mathrm{It} \mathrm{is} \mathrm{an} \mathrm{equation} \mathrm{of} \mathrm{angular} \mathrm{variables} \mathrm{of} \mathrm{motion} \mathrm{corresponding} \mathrm{to} \mathrm{the} \mathrm{linear} \mathrm{variable} \mathrm{based} \mathrm{equations} \mathrm{s}\right] \\ \left(\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{\alpha t}}^2\right) \\ \left(\mathrm{i}\right) \Rightarrow \mathrm{\theta } = 0+\frac{1}{2}{\mathrm{\alpha t}}^2 \\ \mathrm{\theta } =\frac{{\mathrm{\alpha t}}^2}{2}=\frac{2\times 100}{2}=100 \\ \mathrm{n}=\frac{\mathrm{\theta }}{2\mathrm{\pi }}=\frac{100}{2\times 3.14}=15.92 \approx 16 \end{gathered}$$

Velocity of the S wave is v₁ = 4.5 km/s

The velocity of the P waves is v₂ = 8.0 km/s

Let the time taken by the S and P waves to reach the seismograph be t₁ and  t₂

It is given that

    t₁ $-$ t₂ = 4 min

               = 4 × 60 sec

    t₁ $-$ t₂ = 240 sec              .....(i)

Let the distance of the epicentre (km) be S. Then 

    S = v₁t₁ = v₂t₂

⇒  4.5 × t₁ $-$ 8t₂ = 0

⇒              t₁ = $\frac{4.5}{8}$t₁            ....(ii)

Using (i) and (ii)

      t₁ $-$ t₂ = 240

⇒   t₁ $\left(1 - \frac{4.5}{8}\right)$ = 240

⇒     t₁ = $\frac{240 \times 8}{3.5}$

⇒          = 548.5 s

∴      S = v₁t₁ 

           = 4.5 × 548.5

       S = 2468.6 

       S ≈ 2500 km

Velocity is a physical quantity, both magnitude and sirection are needed to define it. 

  S = Vt

S = distance

V = Velocity

t = time

The SI unit of distance is Meter (m)

The SI unit of time is second ( sec)

So the unit of velocity in SI unit is meter per sec i.e  m/s.

Let total distance covered = s

   t₁= $\frac{\left(2/5\right)\mathrm{s}}{{\mathrm{v}}_1}$ and t₂ = $\frac{\left(3/5\right)s}{v_2}$

Average speed = $\frac{\mathrm{total} \mathrm{distance} }{\mathrm{total} \mathrm{time} }$

                   v = $\frac{\mathrm{s}}{{\mathrm{t}}_1+ {\mathrm{t}}_2}$

or                v = $\frac{\mathrm{s}}{{\displaystyle \frac{2\mathrm{s}}{5{\mathrm{v}}_1}+\frac{3\mathrm{s}}{5{\mathrm{v}}_2}}}$

Hence,         v = $\frac{5{\mathrm{v}}_1{\mathrm{v}}_2}{2{\mathrm{v}}_2+3{\mathrm{v}}_1}$

When source is moving towards stationary observer, its apparent frequency

                ${\mathrm{\eta }}_1 = \mathrm{\eta }\left(\frac{\mathrm{\nu }}{\mathrm{\nu } - {\mathrm{v}}_{\mathrm{s}}}\right)$

where, ν$\to$ velocity of sound 

         ν_s $\to$velocity of source

When source is moving away from observer, its apparent frequency

             ${\mathrm{\eta }}_2 = \mathrm{\eta }\left(\frac{\mathrm{\nu }}{\mathrm{\nu } + {\mathrm{v}}_{\mathrm{s}}}\right)$

$\therefore$         $\frac{{\mathrm{\eta }}_1}{{\mathrm{\eta }}_2} = \left(\frac{\mathrm{\nu }}{\mathrm{\nu } - {\mathrm{\nu }}_{\mathrm{s}}}\right) \times \left(\frac{\mathrm{\nu } + {\mathrm{\nu }}_{\mathrm{s}}}{\mathrm{v}}\right)$

             $\frac{5}{3} = \frac{\mathrm{\nu } + {\mathrm{\nu }}_{\mathrm{s}}}{\mathrm{\nu } - {\mathrm{\nu }}_{\mathrm{s}}}$

By solving, 

             ${\mathrm{\nu }}_{\mathrm{s}} = \frac{\mathrm{\nu }}{4} = \frac{340}{4}$

                 = 85 m/s 

During the upward motion, the speed of the body decreases and will be zero at the highest point (since the gravitational force acting downward), afterward, the body starts downward motion and its speed increases.