Question

A person travelling on a straight line moves with a uniform velocity \( v_1 \) for a distance \( x \) and with a uniform velocity \( v_2 \) for the next \( \frac{3x}{2} \) distance. The average velocity in this motion is \( \frac{50}{7} \, \text{m/s} \). If \( v_1 \) is 5 m/s, then \( v_2 \) is ___ m/s.

Hint:

When calculating average velocity in non-uniform motion, break the total distance and time into separate parts. Apply the formula \( v_{\text{avg}} = \frac{total distance}{total time} \) to each segment and solve for the unknown variable.

Answer 1

Correct Answer: 10

Given: \[ v_{\text{avg}} = \frac{x_1 + x_2}{t_1 + t_2} \] Where \( x_1 = x \), \( x_2 = \frac{3x}{2} \), \( v_1 = 5 \, \text{m/s} \), and \( v_2 \) is the unknown velocity. Substituting the values: \[ v_{\text{avg}} = \frac{50}{7} \, \text{m/s} \] \[ \Rightarrow \frac{50}{7} = \frac{x + \frac{3x}{2}}{\frac{x}{v_1} + \frac{3x}{2v_2}} \] \[ \Rightarrow \frac{50}{7} = \frac{\frac{5x}{2}}{\frac{x}{5} + \frac{3x}{2v_2}} \] Simplifying the equation: \[ \Rightarrow \frac{50}{7} = \frac{5x}{2} \times \frac{5}{x} \quad \text{(by cross-multiplying)} \] \[ \Rightarrow v_2 = 10 \, \text{m/s} \]

Answer 2

We need to find the uniform velocity \( v_2 \) for the second part of a journey, given the distances traveled, the velocity for the first part \( v_1 \), and the average velocity for the entire journey.

Concept Used:

The average velocity (\( v_{\text{avg}} \)) is defined as the total displacement divided by the total time taken. For motion in a straight line without a change in direction, the displacement is equal to the total distance.

\[ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} \]

The total time is the sum of the time taken for each segment of the journey, where time \( t \) is calculated as \( t = \frac{\text{distance}}{\text{velocity}} \).

Step-by-Step Solution:

Step 1: Define the total distance traveled.

The person travels a distance \( d_1 = x \) with velocity \( v_1 \) and a distance \( d_2 = \frac{3x}{2} \) with velocity \( v_2 \).

The total distance is:

\[ D_{\text{total}} = d_1 + d_2 = x + \frac{3x}{2} = \frac{2x + 3x}{2} = \frac{5x}{2} \]

Step 2: Define the total time taken.

The time taken for the first part of the journey is \( t_1 = \frac{d_1}{v_1} = \frac{x}{v_1} \).

The time taken for the second part of the journey is \( t_2 = \frac{d_2}{v_2} = \frac{3x/2}{v_2} = \frac{3x}{2v_2} \).

The total time is:

\[ T_{\text{total}} = t_1 + t_2 = \frac{x}{v_1} + \frac{3x}{2v_2} \]

Step 3: Set up the equation for the average velocity.

\[ v_{\text{avg}} = \frac{D_{\text{total}}}{T_{\text{total}}} = \frac{\frac{5x}{2}}{\frac{x}{v_1} + \frac{3x}{2v_2}} \]

Step 4: Simplify the expression for average velocity.

We can factor out \( x \) from the denominator:

\[ v_{\text{avg}} = \frac{\frac{5x}{2}}{x \left( \frac{1}{v_1} + \frac{3}{2v_2} \right)} = \frac{\frac{5}{2}}{\frac{1}{v_1} + \frac{3}{2v_2}} \]

Now, find a common denominator for the terms in the bottom part:

\[ v_{\text{avg}} = \frac{\frac{5}{2}}{\frac{2v_2 + 3v_1}{2v_1v_2}} = \frac{5}{2} \times \frac{2v_1v_2}{3v_1 + 2v_2} = \frac{5v_1v_2}{3v_1 + 2v_2} \]

Step 5: Substitute the given values into the equation.

We are given \( v_{\text{avg}} = \frac{50}{7} \, \text{m/s} \) and \( v_1 = 5 \, \text{m/s} \).

\[ \frac{50}{7} = \frac{5(5)v_2}{3(5) + 2v_2} \] \[ \frac{50}{7} = \frac{25v_2}{15 + 2v_2} \]

Step 6: Solve the equation for \( v_2 \).

Divide both sides by 25:

\[ \frac{50}{7 \times 25} = \frac{v_2}{15 + 2v_2} \] \[ \frac{2}{7} = \frac{v_2}{15 + 2v_2} \]

Now, cross-multiply:

\[ 2(15 + 2v_2) = 7v_2 \] \[ 30 + 4v_2 = 7v_2 \] \[ 30 = 7v_2 - 4v_2 \] \[ 30 = 3v_2 \] \[ v_2 = \frac{30}{3} = 10 \]

The value of \( v_2 \) is 10 m/s.