Motion in Straight Line

Relative velocity is the vector difference between the velocities of two bodies, the velocity of a body with respect to another regarded as being at rest.

Here, we calculate the relative velocity of the jeep:

 153 -45 =108 km/h

$$\begin{gathered} \Rightarrow k\mathrm{m}/\mathrm{hour} =\frac{1000}{60\times 60}=\frac{5}{18} \\ = 108 \times \frac{5}{18} \\ =30 \mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, striking speed = relative speed of a bullet with respect to the thief's car:

 180 - 30 = 150 m/s 

Here :- Initial velocity u=-4.9m/s (-ve is due to vertically upward motion)

Total time t = 2sec

The height of the bridge is given by

s =ut+$\frac{1}{2}{\mathrm{at}}^2$

Here we consider a=g$\Rightarrow$acceleration due to gravity

$$\begin{gathered} \mathrm{s}=-4.9\times 2+\frac{1}{2}\times 9.8\times {\left(2\right)}^2 \\ =-9.8+19.6 =9.8\mathrm{m} \end{gathered}$$

Here:- Mass of the ball m= 10 g ; time dt =0.1 sec ;

acceleration of the ball = 20m/s²

Force F =ma =0.15×20 =3N

Impulsive forces are forces that act over short times producing rapid changes in motion

$\Rightarrow$ Impulsive force = F×dt = 3×0.1 =0.3 Ns

When elevator moves downward with constant acceleration. 

     $\mathrm{\alpha }$ = g.

Then person standing in an elevator find himself weightless.

(g' with respect to the lift is   $\left(\mathrm{g} - \mathrm{\alpha }\right)$

Velocity is rate of change of distance or displacement.
Distance travelled by the particle is
                      
We know that, velocity is rate of change of distance i.e., 


but final velocity v = 0

Hence, distance travelled by the particle before coming to rest is given by
                   

            

Parasec (pc) is an astronomical unit of length equal to the distance at which baseline of one astronomical unit subtends at an angle of one sec of arc.

1 parsec = 3.085677  × 10¹⁶ m = 3.26 light years

Force of attraction between two charges is given by

     F = $\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}} \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

        = $\frac{{\mathrm{A}}^2 {\mathrm{L}}^2}{\mathrm{M} \mathrm{L} {\mathrm{T}}^{-2} {\mathrm{A}}^2}$

       ( where 4$\mathrm{\pi }$ is dimensionless )

    F = [ M^-1 L^-3 T⁴ A² ]

If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
  Work done by the force  = Force x Displacement
or          W = F x s                            ...(i)
But from Newton's 2nd law, we have
  Force  =  Mass x Acceleration
i.e.,       F = ma                                ...(ii)
Hence, from Eqs. (i) and (ii), we get

Hence, Eq (ii) becomes
                
                   
We have given
                   
 

The relative velocity cf thief's jeep with respect to the police jeep

= 153 - 45 = 108 km/h

= $108 \times \frac{5}{18}$ = 30 m/s

Therefore, striking speed = relative speed of bullet with respect to the thief's car

= 180 - 30 = 150 m/s

According to the Kinematic equation

    v² = u² + 2 $\mathrm{\alpha }$s         

 ∴   u² ∝ s  or  ${\mathrm{v}}_{\mathrm{o}}^{ 2}$ ∝ s      ( since  u = v_o )     ....(i)

For given condition

     u'² ∝ 3 s                                      .....(ii)

From eqn. (i) and (ii)

      $\frac{\mathrm{u}{\text{'}}^2}{{\mathrm{v}}_{\mathrm{o}}^{ 2}} = 3$

⇒   u' = $\sqrt{3}$v_o