Set 7k

Q1 [3 Marks]: Define relative velocity in one dimension. If two bodies A and B are moving with velocities $v_A$ and $v_B$ in the same direction, what is the velocity of A relative to B?

Part A: Definition
Relative velocity in one dimension is defined as the time rate of change of relative position of one object with respect to another object. In simple terms, it is the velocity of an object as observed by someone on another moving object.

Part B: Formula
If bodies A and B are moving in the same direction along a straight line, the velocity of A relative to B ($v_{AB}$) is the difference between their individual ground velocities.

$$v_{AB} = v_A - v_B$$

Q2 [3 Marks]: Two trains 120 m and 80 m long are running in opposite directions with velocities 42 km/h and 30 km/h. In what time will they completely cross each other?

Step 1: Convert velocities into SI units (m/s).
Velocity of train 1, $v_1 = 42 \times \frac{5}{18} = \frac{210}{18} = \frac{35}{3}\text{ m/s}$
Velocity of train 2, $v_2 = 30 \times \frac{5}{18} = \frac{150}{18} = \frac{25}{3}\text{ m/s}$

Step 2: Find the relative velocity.
Since the trains are moving in opposite directions, their relative velocity ($v_{rel}$) is the sum of their individual speeds.

$$v_{rel} = v_1 + v_2 = \frac{35}{3} + \frac{25}{3} = \frac{60}{3}$$

$$v_{rel} = 20\text{ m/s}$$

Step 3: Determine the total distance to be covered.
To completely cross each other, the total relative distance covered must equal the sum of the lengths of both trains.

$$\text{Total Distance } (D) = 120\text{ m} + 80\text{ m} = 200\text{ m}$$

Step 4: Calculate the time taken.

$$\text{Time } (t) = \frac{\text{Total Distance}}{\text{Relative Velocity}} = \frac{200}{20}$$

$$t = 10\text{ s}$$

Q3 [4 Marks]: A passenger train of length 60 m travels at 80 m/s. It overtakes a freight train of length 120 m traveling in the same direction at 30 m/s. Calculate the time taken for the passenger train to completely overtake the freight train.

Step 1: Find the relative velocity.
Since both trains are moving in the same direction, the relative velocity of the passenger train with respect to the freight train is the difference between their speeds.

$$v_{rel} = v_{passenger} - v_{freight}$$

$$v_{rel} = 80 - 30 = 50\text{ m/s}$$

Step 2: Determine the total relative distance.
For one train to completely overtake another, the front of the faster train must pass the back of the slower train, and then continue until its back passes the front of the slower train. Therefore, the total relative distance is the sum of their lengths.

$$D = L_{passenger} + L_{freight} = 60 + 120 = 180\text{ m}$$

Step 3: Calculate the time taken.

$$t = \frac{D}{v_{rel}} = \frac{180}{50}$$

$$t = \frac{18}{5} = 3.6$$

$$\text{Time taken } = 3.6\text{ s}$$

Q4 [5 Marks]: On a straight two-lane road, car A is traveling with a speed of 36 km/h. Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC (both 1 km), B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?

Step 1: Convert all velocities to SI units (m/s).

• Velocity of Car A, $v_A = 36 \times \frac{5}{18} = 10\text{ m/s}$
• Velocity of Car B, $v_B = 54 \times \frac{5}{18} = 15\text{ m/s}$
• Velocity of Car C, $v_C = 54 \times \frac{5}{18} = 15\text{ m/s}$

Step 2: Analyze the motion of Car C relative to Car A.
Car A and Car C are approaching each other from opposite directions. Therefore, their relative velocity ($v_{CA}$) is added.

$$v_{CA} = v_C + v_A = 15 + 10 = 25\text{ m/s}$$

The initial distance between A and C is 1 km (**1000 m**). The time ($t$) it takes for C to just reach A is:

$$t = \frac{\text{Distance } AC}{v_{CA}} = \frac{1000}{25} = 40\text{ s}$$

This means Car B has exactly 40 seconds to completely overtake Car A to avoid a collision with Car C.

Step 3: Analyze the motion of Car B relative to Car A.
Car B is behind Car A and moving in the same direction. Their initial relative velocity ($u_{BA}$) is:

$$u_{BA} = v_B - v_A = 15 - 10 = 5\text{ m/s}$$

The initial relative distance B needs to cover to overtake A is 1 km (**1000 m**).
Let $a$ be the required acceleration of Car B.

Step 4: Apply the kinematic equation for relative motion.
We use the equation $S = ut + \frac{1}{2}at^2$ applied to the relative frame of Car A.

$$S_{BA} = u_{BA}t + \frac{1}{2}at^2$$

Substitute the known values ($S = 1000\text{ m}$, $u = 5\text{ m/s}$, $t = 40\text{ s}$):

$$1000 = (5 \times 40) + \frac{1}{2} a (40)^2$$

$$1000 = 200 + \frac{1}{2} a (1600)$$

$$1000 = 200 + 800a$$

Step 5: Solve for minimum acceleration ($a$).

$$800 = 800a$$

$$a = 1$$

$$\text{Minimum acceleration of Car B } = 1\text{ m/s}^2$$