Problem Setup
Let \( d_f \) be the distance travelled on foot and \( d_b \) be the distance travelled on bicycle. Let \( t_f \) be the time spent travelling on foot and \( t_b \) be the time spent travelling on bicycle.
We are given:
- Total distance: \( d_f + d_b = 28 \) km
- Total time: \( t_f + t_b = 8 \) h
- Speed on foot: \( v_f = 2 \) km/h
- Speed on bicycle: \( v_b = 6 \) km/h
Solving for Distance on Foot
Using the relationship distance = speed × time, we have:
- \( d_f = v_f \times t_f \implies t_f = \frac{d_f}{2} \)
- \( d_b = v_b \times t_b \implies t_b = \frac{d_b}{6} \)
Substitute these time expressions into the total time equation:
\( \frac{d_f}{2} + \frac{d_b}{6} = 8 \)
From the total distance equation, we can express \( d_b \) in terms of \( d_f \):
\( d_b = 28 - d_f \)
Substitute this expression for \( d_b \) into the time equation:
\( \frac{d_f}{2} + \frac{28 - d_f}{6} = 8 \)
To solve for \( d_f \), multiply the entire equation by 6 (the least common multiple of 2 and 6):
\( 6 \times \left( \frac{d_f}{2} \right) + 6 \times \left( \frac{28 - d_f}{6} \right) = 6 \times 8 \)
\( 3d_f + (28 - d_f) = 48 \)
Simplify and solve for \( d_f \):
\( 3d_f + 28 - d_f = 48 \)
\( 2d_f + 28 = 48 \)
\( 2d_f = 48 - 28 \)
\( 2d_f = 20 \)
\( d_f = \frac{20}{2} \)
\( d_f = 10 \) km
The distance travelled on foot is 10 km.
