Speedometer shows speed and odometer shows distance travelled, both relative to the surface on which

Q: Why do we add the belt's speed to the car's speed?

Because the car is moving on a surface that is itself moving. From the ground's perspective, the car's total velocity is its velocity relative to the belt plus the belt's velocity relative to the ground.

Q: Why does the odometer measure relative distance?

An odometer works by counting the number of wheel rotations. It only knows how much the wheels have rolled over the surface they are touching, completely ignoring if the surface itself is moving.

Q: How did we get $v_2 t_2 = u_1 t_1$?

We know the absolute distance on belt 1 is $L = (v_1 + u_1)t_1 = v_1 t_1 + u_1 t_1$. We also know the total relative distance is $L = v_1 t_1 + v_2 t_2$. Equating the two gives $v_1 t_1 + u_1 t_1 = v_1 t_1 + v_2 t_2$, which simplifies to $u_1 t_1 = v_2 t_2$.

Q: Why convert time to hours?

Since the speeds are given in km/h and the distance in km, it is mathematically cleaner to convert the total time $\tau = 72$ s into hours ($72/3600 = 0.02$ h) to maintain consistent units throughout the calculation.

Q: What if the car was moving in the opposite direction to the belts?

If the car moved opposite to the belts, its absolute speed would be $v - u$. However, as proven in the solution, this assumption leads to a mathematical contradiction ($u_1 t_1 + v_2 t_2 = 0$), confirming the car must move in the same direction as the belts.

Question

Speedometer shows speed and odometer shows distance travelled, both relative to the surface on which the vehicle moves. Two conveyor-belts each of length $L = 500	ext{ m}$ are arranged along a line one after the other with a negligible gap. The belts are running in the same but unknown direction with constant speeds $u_1 = 20	ext{ km/h}$ and $u_2 = 30	ext{ km/h}$. A toy car installed with both the instruments runs on the belts one after the other spending $	au = 72	ext{ s}$ on them. The speedometer shows constant readings on each of the belts and the odometer shows a total reading of $s = L = 500	ext{ m}$. Find the speedometer readings on each of the belts.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a moving walkway at an airport. If you walk forward, your speed relative to the ground is the sum of your walking speed and the walkway's speed. This problem presents a similar scenario but with a brilliant twist involving the instruments of a toy car.

The car is equipped with a speedometer and an odometer. It is crucial to understand what these instruments actually measure. A speedometer measures the speed of the wheels rotating against the surface they are touching. Therefore, it measures the **relative speed** of the car with respect to the belt. Let's call these relative speeds $v_1$ and $v_2$ on the first and second belts, respectively.

Similarly, the odometer counts the total number of wheel rotations, meaning it measures the **total relative distance** covered by the car on the belts. The problem states that the total odometer reading is $s = L = 500 	ext{ m}$. This gives us our first fundamental constraint:

$$v_1 t_1 + v_2 t_2 = L$$

### The Direction Deduction

Before we dive into the algebra, we must address a hidden ambiguity in the problem: Are the belts moving in the same direction as the car, or in the opposite direction?

Let's perform a quick thought experiment. What if the belts were moving opposite to the car? To cross the belt of length $L$, the car's absolute speed would be $v_1 - u_1$. The distance covered on the ground would be $(v_1 - u_1)t_1 = L$. Expanding this gives $v_1 t_1 = L + u_1 t_1$.

Now, look back at our odometer constraint: $v_1 t_1 + v_2 t_2 = L$. If we substitute our expanded term, we get $L + u_1 t_1 + v_2 t_2 = L$, which simplifies to $u_1 t_1 + v_2 t_2 = 0$. Since time and speed are strictly positive quantities, this equation is mathematically impossible!

This elegant contradiction proves that the car **must** be moving in the exact same direction as the conveyor belts.

### The Master Equation

Now that we have established the direction of motion, we can set up our kinematic equations for absolute motion. From the ground's perspective, the car covers a distance $L$ on each belt.

On the first belt, the absolute speed is $v_1 + u_1$. Therefore, the distance equation is:

$$(v_1 + u_1)t_1 = L \implies v_1 t_1 + u_1 t_1 = L$$

On the second belt, the absolute speed is $v_2 + u_2$. The distance equation is:

$$(v_2 + u_2)t_2 = L \implies v_2 t_2 + u_2 t_2 = L$$

Here comes the master stroke. We know from the odometer that $v_1 t_1 + v_2 t_2 = L$. But our first belt equation tells us that $v_1 t_1 + u_1 t_1$ is also equal to $L$. By equating these two expressions, we find that $v_2 t_2$ must perfectly equal $u_1 t_1$.

Substituting $v_2 t_2 = u_1 t_1$ into our second belt equation yields a beautiful, simplified relation that completely eliminates the unknown relative speeds:

$$u_1 t_1 + u_2 t_2 = L$$

### Final Calculation

We now have a clean system of two linear equations involving only the time intervals $t_1$ and $t_2$:

1. $t_1 + t_2 = 	au$
2. $u_1 t_1 + u_2 t_2 = L$

Solving this system simultaneously gives us the exact time spent on each belt:

$$t_1 = \frac{L - u_2 	au}{u_1 - u_2}$$
$$t_2 = \frac{u_1 	au - L}{u_1 - u_2}$$

To maintain consistent units, we convert the total time $	au = 72 	ext{ s}$ into hours: $	au = \frac{72}{3600} = 0.02 	ext{ h}$. The length $L$ is $0.5 	ext{ km}$. Plugging in the given speeds $u_1 = 20 	ext{ km/h}$ and $u_2 = 30 	ext{ km/h}$:

$$t_1 = \frac{0.5 - 30(0.02)}{20 - 30} = \frac{0.5 - 0.6}{-10} = 0.01 	ext{ h}$$
$$t_2 = \frac{20(0.02) - 0.5}{20 - 30} = \frac{0.4 - 0.5}{-10} = 0.01 	ext{ h}$$

The car spends exactly $0.01 	ext{ hours}$ on each belt! Finally, we can back-substitute these time values to find the speedometer readings:

$$v_1 = \frac{L}{t_1} - u_1 = \frac{0.5}{0.01} - 20 = 50 - 20 = 30 	ext{ km/h}$$
$$v_2 = \frac{L}{t_2} - u_2 = \frac{0.5}{0.01} - 30 = 50 - 30 = 20 	ext{ km/h}$$

The speedometer reads $30 	ext{ km/h}$ on the first belt and $20 	ext{ km/h}$ on the second belt. A perfect harmony of relative kinematics!

Visualized Solution

Understanding the Setup

Instrument Readings

Direction of Motion Deduction

Absolute Motion Equations

Utilizing the Odometer Constraint

Solving for Time Intervals $t_{1}$ and $t_{2}$

Calculating Time Values

Final Speedometer Readings $v_{1}$ and $v_{2}$

The Sigma Insight: Relative Velocity

Analyzing the Setup

The Direction Deduction

The Master Equation

Final Calculation

Similar Questions

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time $t_{1}$ . If he remains stationary on a moving escalator, then the escalator takes him up in time $t_{2}$ . The time taken by him to walk up on the moving escalator will be