Numerical Problems Based on Relative Velocity in 1D for Class 11 Physics

Let the direction of the jet airplane be the positive direction.

Velocity of the jet w.r.t. earth, $v_J = 450\text{ kmh}^{-1}$.

The gases are ejected backwards (opposite to the jet), so the velocity of the gases w.r.t. the jet is $v_{GJ} = -1200\text{ kmh}^{-1}$.

Using the relative velocity formula: $v_{GJ} = v_G – v_J$, where $v_G$ is the velocity of the gases w.r.t. the earth.

The negative sign indicates that the gases are travelling in the direction opposite to the jet. The speed (magnitude) is $750\text{ kmh}^{-1}$.

Let the eastward direction be positive.

(i) Both cars travelling eastwards:
$v_A = +60\text{ kmh}^{-1}$, $v_B = +45\text{ kmh}^{-1}$

Relative velocity of A w.r.t. B ($v_{AB}$):

The positive sign means it is $15\text{ kmh}^{-1}$ eastwards.

(ii) Car A travelling eastwards, car B travelling westwards:
$v_A = +60\text{ kmh}^{-1}$, $v_B = -45\text{ kmh}^{-1}$

Relative velocity of A w.r.t. B ($v_{AB}$):

The positive sign means it is $105\text{ kmh}^{-1}$ eastwards.

Let the direction of the car be the positive direction.

Velocity of the car w.r.t. ground, $v_C = 100\text{ kmh}^{-1}$.

The bullet is fired in the opposite direction, so the velocity of the bullet w.r.t. the car is $v_{BC} = -250\text{ kmh}^{-1}$.

Using the relative velocity formula: $v_{BC} = v_B – v_C$, where $v_B$ is the velocity of the bullet w.r.t. the ground.

The negative sign indicates the bullet travels opposite to the car. Its speed (magnitude) is $150\text{ kmh}^{-1}$.

First, convert the time into hours to match the speed units:

Calculate the distance traveled by each car from the starting point:

Distance traveled by car A ($s_A$):

Distance traveled by car B ($s_B$):

To find the position of car A with respect to car B, we calculate the difference:

The negative sign indicates that car A is $5\text{ km}$ behind car B.