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Chapter 2: Problem 11

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=8 \mathrm{~ms}^{-1}\) along a straight horizontal road. \(\mathrm{A}\) pedestrian starts to cross the road with a uniform speed \(v\) when the truck is \(4 \mathrm{~m}\) away from him. The minimum value of \(v\), so that he can cross the road safely is \(\begin{array}{ll}\text { (a) } 2.62 \mathrm{~ms}^{-1} & \text { (b) } 4.6 \mathrm{~ms}^{-1}\end{array}\) (c) \(3.57 \mathrm{~ms}^{-1}\) (d) \(1.414 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The minimum speed for the pedestrian to cross safely is (b) 4.6 ms^-1.

Step by step solution

01

Understand the Problem

The truck is moving towards the pedestrian. To prevent a collision, the pedestrian must cross the 2 meters of road while the truck travels the remaining 4 meters to reach him. We need to find the minimum speed of the pedestrian, \( v \), to achieve this.
02

Calculate Time for Truck

The time taken by the truck, \( T_{truck} \), to reach the pedestrian's starting point is the distance (4 meters) divided by its speed (8 m/s). \[ T_{truck} = \frac{4}{8} = 0.5 \text{ seconds} \]
03

Calculate Required Speed of Pedestrian

The pedestrian must cross 2 meters before the truck arrives. Using \( T_{truck} = 0.5 \) seconds, the required speed \( v \) is given by: \[ v = \frac{2}{T_{truck}} = \frac{2}{0.5} = 4 \text{ ms}^{-1} \]
04

Verify the Provided Options

Among the given options, we check which is closest to the calculated necessary speed. The options given are: (a) 2.62 ms^-1, (b) 4.6 ms^-1, (c) 3.57 ms^-1, (d) 1.414 ms^-1. Since 4 ms^-1 is not precisely an option and observing the objective is to not be lower than this speed, we choose slightly higher, which is the closest available option (b) 4.6 ms^-1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relative Velocity
Relative velocity is a way of describing the velocity of one object as observed from another. In this exercise, the relative velocity concept helps us understand how fast the pedestrian needs to move relative to the truck. Even though both the pedestrian and the truck are in motion, we are specifically interested in their speeds as they move towards each other.
Here's what happens:
  • The truck moves towards the pedestrian at a given velocity.
  • The pedestrian starts moving at a different speed.
To avoid the collision, we need the pedestrian's relative speed to be enough to cross the road before the truck reaches his starting point. By understanding the relative motions, we can effectively calculate the minimum velocity required for the pedestrian.
Uniform Speed
Uniform speed refers to constant speed, where neither acceleration nor deceleration occurs. In this example, both the pedestrian and the truck move at uniform speeds.
Key points about uniform speed:
  • The truck travels at a constant speed of 8 m/s, meaning it covers equal distances in equal intervals of time.
  • The pedestrian's speed also remains constant, which is necessary for simplifying calculations and ensuring accurate predictions.
Uniform speed helps us calculate the time required for each party to travel their respective distances, ensuring we can figure out the pedestrian's cutoff speed to avoid the collision.
Collision Avoidance
Avoiding collisions is crucial, especially when two entities are moving towards each other. The pedestrian and the truck have specific paths and speeds, and resolving how the pedestrian avoids collision is key.
The strategy here involves:
  • The pedestrian needs to start moving as soon as the truck is identified.
  • The calculation centers on ensuring the pedestrian moves fast enough to reach the other side before the truck gets to his initial position.
By determining the minimum safe speed, the pedestrian can cross without risking an accident. The application of physics helps ensure these calculations are precise and reliable.
Distance-Time Relationship
The distance-time relationship is a fundamental concept that helps us understand motion through the equations of motion. Here鈥檚 how it applies:
  • The truck moves 4 meters toward the pedestrian, taking 0.5 seconds at 8 m/s.
  • The pedestrian needs to cross 2 meters in this 0.5-second window.
Thus, the pedestrian's required speed is calculated using this relationship: \[ v = \frac{2 \, \text{meters}}{0.5 \, \text{seconds}} = 4 \, \text{m/s} \]
Calculating how much distance each moving entity covers in time allows for precise planning to avoid any intersection, aiding in collision avoidance.

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Most popular questions from this chapter

A particle of mass \(m\) moves on the \(x\)-axis as follows : it starts from rest at \(t=0\) from the point \(x=0\) and comes to rest at \(t=1\) at the point \(x=1\). No other information is available about its motion at intermediate time \((0

The engine of a train can impart a maximum acceleration of \(1 \mathrm{~ms}^{-2}\) and the brakes can give a maximum retardation of \(3 \mathrm{~ms}^{-2} .\) The least time during which a train can go from one place to the other place at a distance of \(1.2 \mathrm{~km}\) is nearly (a) \(108 \leq\) (b) \(191 \mathrm{~s}\) (c) \(56.6 \mathrm{~s}\) (d) time is fixed

The motion of a body is given by the equation \(\frac{d v(t)}{d t}=6.0-3 v(t)\), where \(v(t)\) is speed in \(\mathrm{ms}^{-1}\) and \(t\) in second. If body was at rest at \(t=0\) (a) the terminal speed is \(2.0 \mathrm{~ms}^{-1}\) (b) the speed varies with the times as \(v(t)=2\left(1-e^{-3 t}\right) \mathrm{ms}^{-1}\) (c) the speed is \(1.0 \mathrm{~ms}^{-1}\) when the acceleration is half the initial value (d) the magnitude of the initial acceleration is \(6.0 \mathrm{~ms}^{-2}\)

The driver of a car moving with a speed of \(10 \mathrm{~ms}^{-1}\) sees a red light ahead, applies brakes and stops after covering \(10 \mathrm{~m}\) distance. If the same car were moving with a speed of \(20 \mathrm{~ms}^{-1}\), the same driver would have stopped the car after covering \(30 \mathrm{~m}\) distance. Within what distance the car can be stopped if travelling with a velocity of \(15 \mathrm{~ms}^{-1}\) ? Assume the same reaction time and the same deceleration in each case. (a) \(18.75 \mathrm{~m}\) (b) \(20.75 \mathrm{~m}\) (c) \(22.75 \mathrm{~m}\) (d) \(25 \mathrm{~m}\)

A particle moving in a straight line with uniform acceleration is observed to be a distance \(a\) from a fixed point initially. It is at distances \(b, c, d\) from the same point after \(n, 2 n, 3 n\) second. The acceleration of the particle is (a) \(\frac{c-2 b+a}{n^{2}}\) (b) \(\frac{c+b+a}{9 n^{2}}\) (c) \(\frac{c+2 b+a}{4 n^{2}}\) (d) \(\frac{c-b+a}{n^{2}}\)
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