Pathfinder for Olympiad and JEE Advanced Physics

LEVELJEE Advanced

Animated Solution for Physics - Kinematics: On a straight section of a highway, sensors are installed to measure traffic density. For each lane of the highway, a sensor runs on an overhead wire. On a particular day a sensor running with a speed $u = 5 km/h$ opposite to the flow of traffic underneath, counts $N = 360$ vehicles in a length $L = 1 km$ of the highway. If all the vehicles are moving with the same constant speed $v = 40 km/h$ and density of the vehicles is uniform, calculate number of vehicles per $l = 100 m$ of the lane.

Q: Does the length $L$ refer to the distance on the ground or the distance relative to the cars?

The problem states the sensor counts vehicles 'in a length $L=1$ km of the highway', which means $L$ is the physical distance measured along the ground.

Q: Why can't we just use $N/L$ to find the density?

If the sensor were stationary, $N/L$ would not be the density either, it would depend on time. If you took a snapshot, $N/L$ is density. But the sensor takes time to cover $L$, and during this time, more cars drive into the segment.

Q: What if the sensor was moving in the same direction as the traffic?

Then the relative velocity would be $v - u$ (assuming $v > u$). The sensor would count fewer cars because the cars are running away from it.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Traffic Flow}

Let the linear density of vehicles be $λ$ (vehicles per unit length).
The sensor moves with speed $u$ opposite to the traffic moving with speed $v$ .

\text{Relative Velocity of Traffic}

Relative velocity of the traffic with respect to the sensor:
$v_{rel} = v - (- u) = v + u$

\text{Time Taken by Sensor}

The sensor covers a distance $L$ on the highway.
Time taken by the sensor:
$t = \frac{L}{u}$

\text{Effective Distance Scanned}

Distance covered by the traffic relative to the sensor in time $t$ :
$d_{rel} = v_{rel} \times t$
$d_{rel} = (v + u) (\frac{L}{u})$

\text{Total Vehicles Counted}

Total number of vehicles counted ( $N$ ) is the density times the relative distance:
$N = λ \times d_{rel}$
$N = λ \frac{L ( v + u )}{u}$

\text{Linear Density of Vehicles}

Rearranging for linear density $λ$ :
$λ = \frac{N u}{L ( v + u )}$

\text{Vehicles in Length } l

Number of vehicles in length $l$ is $n = λ l$ :
$n = \frac{N u l}{L ( v + u )}$
Substitute $N = 360$ , $u = 5 km/h$ , $v = 40 km/h$ , $L = 1 km$ , $l = 0.1 km$ :
$n = \frac{360 \times 5 \times 0.1}{1 \times ( 40 + 5 )} = \frac{180}{45} = 4$

\text{The Way Forward}

What if the sensor moved in the same direction as the traffic?
The relative velocity would be $v - u$ .
The counted vehicles would be $N^{'} = λ \frac{L ( v - u )}{u}$ .

00:00 / 00:00

The Sigma Insight: Relative Velocity

Solution Diagram

The Illusion of Speed

Imagine standing on the side of a highway, watching cars zoom by. Now, imagine you are running towards those cars. They seem to be moving much faster, right? This is the core of relative velocity, and it's the secret to unlocking this problem.

The sensor isn't just sitting there; it's actively moving against the flow of traffic. This means it sweeps through the cars faster than if it were stationary.

Setting Up the Variables

Let's define what we know. The traffic is moving at a constant speed

v = 40 km/h

. The sensor is moving in the opposite direction at

u = 5 km/h

Because they are moving towards each other, the relative velocity of the traffic with respect to the sensor is the sum of their speeds:

v_{rel} = v + u

The Effective Distance

The sensor covers a physical distance

L = 1 km

along the highway. How long does this take? Time is simply distance divided by speed.

t = \frac{L}{u}

During this time

t

, how much "length" of traffic has actually passed under the sensor? We call this the relative distance,

d_{rel}

d_{rel} = v_{rel} \times t = (v + u) (\frac{L}{u})

Finding the Density

The sensor counts

N = 360

vehicles in this effective distance. If we assume the cars are evenly spaced, we can define a linear density

λ

, which is the number of cars per unit length.

The total number of cars

N

is simply the density multiplied by the effective distance:

N = λ \times d_{rel} = λ \frac{L ( v + u )}{u}

Rearranging this to solve for our unknown density

λ

λ = \frac{N u}{L ( v + u )}

The Final Calculation

We don't just want the density; we want the number of vehicles

n

in a specific length

l = 100 m

(which is

0.1 km

We multiply the density by this length:

n = λ l = \frac{N u l}{L ( v + u )}

Now, let's plug in the numbers. Notice how keeping all our units in kilometers and hours prevents any messy conversions!

n = \frac{360 \times 5 \times 0.1}{1 \times ( 40 + 5 )}

n = \frac{180}{45} = 4

There are exactly 4 vehicles in every 100 meters of the lane. A perfectly clean integer!

Common Traps & FAQs

Why do we add the speeds

u

and

v

Because the sensor and the cars are moving in opposite directions. From the sensor's frame of reference, the cars are approaching faster than their actual ground speed.

Does the length

L

refer to the distance on the ground or the distance relative to the cars?

The problem states the sensor counts vehicles 'in a length

L = 1

km of the highway', which means

L

is the physical distance measured along the ground.

Why can't we just use

N / L

to find the density?

If the sensor were stationary,

N / L

would not be the density either, it would depend on time. If you took a snapshot,

N / L

is density. But the sensor takes time to cover

L

, and during this time, more cars drive into the segment.

What if the sensor was moving in the same direction as the traffic?

Then the relative velocity would be

v - u

(assuming

v > u

). The sensor would count fewer cars because the cars are running away from it.

Do I need to convert km/h to m/s?

Not necessarily! As long as you keep your units consistent (e.g., using km for distance and km/h for speed), the time units will cancel out perfectly. Just remember to convert

l = 100

m to

0.1

km.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Traffic Flow}

\text{Relative Velocity of Traffic}

\text{Time Taken by Sensor}

\text{Effective Distance Scanned}

\text{Total Vehicles Counted}

\text{Linear Density of Vehicles}

\text{Vehicles in Length } l

\text{The Way Forward}

The Sigma Insight: Relative Velocity

The Illusion of Speed

Setting Up the Variables

The Effective Distance

Finding the Density

The Final Calculation

Similar Questions

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Traffic Flow}

\text{Relative Velocity of Traffic}

\text{Time Taken by Sensor}

\text{Effective Distance Scanned}

\text{Total Vehicles Counted}

\text{Linear Density of Vehicles}

\text{Vehicles in Length } l

\text{The Way Forward}

The Sigma Insight: Relative Velocity

The Illusion of Speed

Setting Up the Variables

The Effective Distance

Finding the Density

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water in km/h is

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What is speed of the particle C at the instant t=5 s?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What is modulus of acceleration of the particle C at the instant t=5 s?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

A boat which has a speed of $5 km/h$ in still water crosses a river of width $1 km$ along the shortest possible path in $15 min$ . The velocity of the river water in km/h is

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?