Physics Solution: Relative Motion on Two Belts
Problem Statement
A toy car runs on two sequential conveyor belts of length $L=500$ m each. The belts move in the same direction with speeds $u_1 = 20$ km/h and $u_2 = 30$ km/h.
Conditions:
1. Total Odometer reading (relative distance) $s = L = 0.5$ km.
2. Total Time $\tau = 72$ s $= 0.02$ h.
3. Speedometer reading (relative speed) is constant on each belt.
Step 1: Equations of Motion
Let $t_1$ and $t_2$ be the time spent on belt 1 and belt 2 respectively.
Condition 2 implies:
$$ t_1 + t_2 = 0.02 \text{ h} \quad \text{— (1)} $$
On each belt, the car must traverse the physical length $L$. The ground distance covered on belt 1 is $L$, and on belt 2 is $L$.
Relation between ground distance, relative distance, and belt distance:
$$ D_{ground} = D_{relative} + D_{belt} $$
$$ L = d_{r1} + u_1 t_1 $$
$$ L = d_{r2} + u_2 t_2 $$
Summing the relative distances (Condition 1: Total odometer = $L$):
$$ d_{r1} + d_{r2} = L $$
Substitute $d_{r} = L – u t$:
$$ (L – u_1 t_1) + (L – u_2 t_2) = L $$
$$ 2L – (u_1 t_1 + u_2 t_2) = L $$
$$ u_1 t_1 + u_2 t_2 = L \quad \text{— (2)} $$
Step 2: Solving for Time
We have a system of linear equations:
1. $t_1 + t_2 = 0.02$
2. $20 t_1 + 30 t_2 = 0.5$
From (1), $t_1 = 0.02 – t_2$. Substitute into (2):
$$ 20(0.02 – t_2) + 30 t_2 = 0.5 $$
$$ 0.4 – 20 t_2 + 30 t_2 = 0.5 $$
$$ 10 t_2 = 0.1 $$
$$ t_2 = 0.01 \text{ h} $$
$$ t_1 = 0.02 – 0.01 = 0.01 \text{ h} $$
Step 3: Calculating Speedometer Readings
The speedometer measures relative speed $v_r = d_r / t$.
On Belt 1:
$$ d_{r1} = L – u_1 t_1 = 0.5 – 20(0.01) = 0.5 – 0.2 = 0.3 \text{ km} $$
$$ v_{r1} = \frac{0.3}{0.01} = 30 \text{ km/h} $$
On Belt 2:
$$ d_{r2} = L – u_2 t_2 = 0.5 – 30(0.01) = 0.5 – 0.3 = 0.2 \text{ km} $$
$$ v_{r2} = \frac{0.2}{0.01} = 20 \text{ km/h} $$
Answer:
First belt: 30 km/h
Second belt: 20 km/h
Problem Statement
A toy car runs on two sequential conveyor belts of length $L=500$ m each. The belts move in the same direction with speeds $u_1 = 20$ km/h and $u_2 = 30$ km/h.
Conditions:
1. Total Odometer reading (relative distance) $s = L = 0.5$ km.
2. Total Time $\tau = 72$ s $= 0.02$ h.
3. Speedometer reading (relative speed) is constant on each belt.
Step 1: Equations of Motion
Let $t_1$ and $t_2$ be the time spent on belt 1 and belt 2 respectively.
Condition 2 implies:
$$ t_1 + t_2 = 0.02 \text{ h} \quad \text{— (1)} $$
On each belt, the car must traverse the physical length $L$. The ground distance covered on belt 1 is $L$, and on belt 2 is $L$.
Relation between ground distance, relative distance, and belt distance:
$$ D_{ground} = D_{relative} + D_{belt} $$
$$ L = d_{r1} + u_1 t_1 $$
$$ L = d_{r2} + u_2 t_2 $$
Summing the relative distances (Condition 1: Total odometer = $L$): $$ d_{r1} + d_{r2} = L $$ Substitute $d_{r} = L – u t$: $$ (L – u_1 t_1) + (L – u_2 t_2) = L $$ $$ 2L – (u_1 t_1 + u_2 t_2) = L $$ $$ u_1 t_1 + u_2 t_2 = L \quad \text{— (2)} $$
Step 2: Solving for Time
We have a system of linear equations: 1. $t_1 + t_2 = 0.02$ 2. $20 t_1 + 30 t_2 = 0.5$
From (1), $t_1 = 0.02 – t_2$. Substitute into (2): $$ 20(0.02 – t_2) + 30 t_2 = 0.5 $$ $$ 0.4 – 20 t_2 + 30 t_2 = 0.5 $$ $$ 10 t_2 = 0.1 $$ $$ t_2 = 0.01 \text{ h} $$ $$ t_1 = 0.02 – 0.01 = 0.01 \text{ h} $$
Step 3: Calculating Speedometer Readings
The speedometer measures relative speed $v_r = d_r / t$.
On Belt 1:
$$ d_{r1} = L – u_1 t_1 = 0.5 – 20(0.01) = 0.5 – 0.2 = 0.3 \text{ km} $$
$$ v_{r1} = \frac{0.3}{0.01} = 30 \text{ km/h} $$
On Belt 2:
$$ d_{r2} = L – u_2 t_2 = 0.5 – 30(0.01) = 0.5 – 0.3 = 0.2 \text{ km} $$
$$ v_{r2} = \frac{0.2}{0.01} = 20 \text{ km/h} $$
Answer:
First belt: 30 km/h
Second belt: 20 km/h