Physics Solution: Escalator Kinetics
Problem Statement
Two boys climb a moving escalator. Both move relative to the escalator steps with speed $v_r = 50$ cm/s, but with different stepping cycles. We need to find the speed of the escalator $v_e$.
Given Data:
- Relative speed on steps: $v_r = 50$ cm/s
- Boy 1: $p_1 = 1$ step up, $q_1 = 2$ steps down. Time $t_1 = 250$ s.
- Boy 2: $p_2 = 2$ steps up, $q_2 = 1$ step down. Time $t_2 = 50$ s.
Step-by-Step Solution
1. Effective Velocities Relative to Escalator
Let $L_0$ be the length of one step. The time for one full cycle (up and down moves) is determined by the total distance covered on the steps surface divided by the relative speed $v_r$.
For Boy 1:
Distance covered on steps per cycle: $(p_1 + q_1)L_0$.
Net displacement relative to escalator per cycle: $(p_1 – q_1)L_0$.
Time per cycle: $\Delta t = \frac{(p_1+q_1)L_0}{v_r}$.
Average velocity relative to escalator $u_1$:
$$ u_1 = \frac{\text{Net Displacement}}{\text{Time}} = \frac{(p_1 – q_1)L_0}{(p_1+q_1)L_0 / v_r} = v_r \left( \frac{p_1 – q_1}{p_1 + q_1} \right) $$
Substituting values ($p_1=1, q_1=2$):
$$ u_1 = 50 \left( \frac{1 – 2}{1 + 2} \right) = 50 \left( -\frac{1}{3} \right) = -\frac{50}{3} \text{ cm/s} $$
For Boy 2:
Similarly for Boy 2 ($p_2=2, q_2=1$):
$$ u_2 = v_r \left( \frac{p_2 – q_2}{p_2 + q_2} \right) = 50 \left( \frac{2 – 1}{2 + 1} \right) = 50 \left( \frac{1}{3} \right) = \frac{50}{3} \text{ cm/s} $$
2. Velocity Relative to Ground
Let $v_e$ be the escalator speed. The ground velocity $V$ is the sum of the escalator speed and the effective relative velocity.
$$ V_1 = v_e + u_1 $$
$$ V_2 = v_e + u_2 $$
Since both boys cover the same ground height/distance $H$ (from ground to first floor):
$$ H = V_1 t_1 = V_2 t_2 $$
$$ (v_e + u_1) t_1 = (v_e + u_2) t_2 $$
3. Solving for Escalator Speed ($v_e$)
Rearranging the equation:
$$ v_e t_1 + u_1 t_1 = v_e t_2 + u_2 t_2 $$
$$ v_e (t_1 – t_2) = u_2 t_2 – u_1 t_1 $$
$$ v_e = \frac{u_2 t_2 – u_1 t_1}{t_1 – t_2} $$
Substitute the values:
$$ u_1 = -50/3, \quad t_1 = 250 $$
$$ u_2 = 50/3, \quad t_2 = 50 $$
$$ v_e = \frac{ (50/3)(50) – (-50/3)(250) }{ 250 – 50 } $$
$$ v_e = \frac{ \frac{2500}{3} + \frac{12500}{3} }{ 200 } $$
$$ v_e = \frac{ 15000/3 }{ 200 } = \frac{ 5000 }{ 200 } = 25 \text{ cm/s} $$
Answer: The escalator is running at 25 cm/s.
Figure: Velocity vectors relative to the escalator.
Problem Statement
Two boys climb a moving escalator. Both move relative to the escalator steps with speed $v_r = 50$ cm/s, but with different stepping cycles. We need to find the speed of the escalator $v_e$.
Given Data:
- Relative speed on steps: $v_r = 50$ cm/s
- Boy 1: $p_1 = 1$ step up, $q_1 = 2$ steps down. Time $t_1 = 250$ s.
- Boy 2: $p_2 = 2$ steps up, $q_2 = 1$ step down. Time $t_2 = 50$ s.
Step-by-Step Solution
1. Effective Velocities Relative to Escalator
Let $L_0$ be the length of one step. The time for one full cycle (up and down moves) is determined by the total distance covered on the steps surface divided by the relative speed $v_r$.
For Boy 1:
Distance covered on steps per cycle: $(p_1 + q_1)L_0$.
Net displacement relative to escalator per cycle: $(p_1 – q_1)L_0$.
Time per cycle: $\Delta t = \frac{(p_1+q_1)L_0}{v_r}$.
Average velocity relative to escalator $u_1$:
$$ u_1 = \frac{\text{Net Displacement}}{\text{Time}} = \frac{(p_1 – q_1)L_0}{(p_1+q_1)L_0 / v_r} = v_r \left( \frac{p_1 – q_1}{p_1 + q_1} \right) $$
Substituting values ($p_1=1, q_1=2$):
$$ u_1 = 50 \left( \frac{1 – 2}{1 + 2} \right) = 50 \left( -\frac{1}{3} \right) = -\frac{50}{3} \text{ cm/s} $$
For Boy 2:
Similarly for Boy 2 ($p_2=2, q_2=1$):
$$ u_2 = v_r \left( \frac{p_2 – q_2}{p_2 + q_2} \right) = 50 \left( \frac{2 – 1}{2 + 1} \right) = 50 \left( \frac{1}{3} \right) = \frac{50}{3} \text{ cm/s} $$
2. Velocity Relative to Ground
Let $v_e$ be the escalator speed. The ground velocity $V$ is the sum of the escalator speed and the effective relative velocity. $$ V_1 = v_e + u_1 $$ $$ V_2 = v_e + u_2 $$
Since both boys cover the same ground height/distance $H$ (from ground to first floor): $$ H = V_1 t_1 = V_2 t_2 $$ $$ (v_e + u_1) t_1 = (v_e + u_2) t_2 $$
3. Solving for Escalator Speed ($v_e$)
Rearranging the equation: $$ v_e t_1 + u_1 t_1 = v_e t_2 + u_2 t_2 $$ $$ v_e (t_1 – t_2) = u_2 t_2 – u_1 t_1 $$ $$ v_e = \frac{u_2 t_2 – u_1 t_1}{t_1 – t_2} $$
Substitute the values: $$ u_1 = -50/3, \quad t_1 = 250 $$ $$ u_2 = 50/3, \quad t_2 = 50 $$ $$ v_e = \frac{ (50/3)(50) – (-50/3)(250) }{ 250 – 50 } $$ $$ v_e = \frac{ \frac{2500}{3} + \frac{12500}{3} }{ 200 } $$ $$ v_e = \frac{ 15000/3 }{ 200 } = \frac{ 5000 }{ 200 } = 25 \text{ cm/s} $$
Answer: The escalator is running at 25 cm/s.
Figure: Velocity vectors relative to the escalator.