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A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=8 \mathrm{~m} / \mathrm{s}\) along a straight horizontal road. 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 a. \(2.62 \mathrm{~m} / \mathrm{s}\) b. \(4.6 \mathrm{~m} / \mathrm{s}\) c. \(3.57 \mathrm{~m} / \mathrm{s}\) d. \(1.414 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The minimum speed is 4 m/s; hence the closest option is b: 4.6 m/s.

Step by step solution

01

Understanding the Scenario

We need to calculate the minimum speed at which the pedestrian can walk so that he can cross a 2 m wide road safely without being hit by a truck coming at him at a speed of 8 m/s, and when he starts crossing from a point that is 4 m away from the approaching truck.
02

Calculating the Time Available to Cross

The pedestrian must cross before the truck reaches his starting position. Since the truck is 4 m away and moving at 8 m/s, the time the pedestrian has to clear the road is \( t = \frac{4}{8} = 0.5 \text{ seconds} \).
03

Determining the Minimum Speed

To cross the 2 m road safely within 0.5 seconds, calculate the pedestrian’s minimum speed using the formula: \( v = \frac{d}{t} \), where \( d = 2 \text{ meters} \) and \( t = 0.5 \text{ seconds} \). Thus, \( v = \frac{2}{0.5} = 4 \text{ m/s} \).
04

Comparing with Given Options

Review the options to pick the closest or matching speed. Option b, \( 4.6 \text{ m/s} \), is slightly above our calculated minimum of 4 m/s which ensures safety against any calculation margin errors, hence it is the best choice.

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Key Concepts

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

Uniform Speed
Uniform speed means moving at a constant rate without accelerating or decelerating. When something moves at uniform speed, it covers equal distances in equal intervals of time. For instance, if a truck travels at a uniform speed of 8 m/s, this means that every second, it travels 8 meters. This concept is crucial in problems involving motion as it allows for simple calculation of distance, time, and speed using the formula:\[ \text{Distance} = \text{Speed} \times \text{Time} \]If you know two of these quantities, you can find the third. In our problem, the truck’s uniform speed helps determine how long the pedestrian has to cross before it reaches his starting point. Once you understand uniform speed, solving related motion problems becomes more straightforward.
Relative Motion
Relative motion helps us understand how different objects in motion relate to each other. Imagine standing on a road and watching a truck approach at 8 m/s. If you are a pedestrian about to cross, you need to think about not only how fast you are moving but also how your speed relates to the truck’s speed. Relative motion is the comparison of different speeds between objects. In this scenario, as the truck moves towards you, you need to move at a certain minimum speed away from its path to avoid being hit. The time available for crossing the road can be thought of in terms of the truck 'closing in' on the starting point. By understanding the truck’s velocity relative to your own speed, you can calculate how quickly you need to move to stay safe. Always consider the idea of relative speed, especially in scenarios involving two moving objects.
Problem Solving
Problem solving in physics is a systematic process. Understanding the problem thoroughly is always the first step. In our scenario, we need to calculate the minimum safe speed for the pedestrian.Begin by identifying what you know:
  • The width of the road is 2 meters.
  • The truck is 4 meters away and moves at 8 m/s.
Next, calculate the time available until the truck reaches the starting point by using the truck’s speed:\[ t = \frac{\text{Distance to truck}}{\text{Truck's speed}} = \frac{4}{8} = 0.5 \text{ seconds} \]Now use this time to find the speed needed by the pedestrian to safely cross the road's width of 2 meters:\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{2}{0.5} = 4 \text{ m/s} \]Choose the answer from the options that is closest or slightly higher to ensure the pedestrian's safety with some margin. Step-by-step analysis is crucial in understanding complex physics problems and finding accurate solutions.

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Most popular questions from this chapter

Two cars are moving in same direction with a speed of \(30 \mathrm{~km} / \mathrm{h}\). They are separated by a distance of \(5 \mathrm{~km}\). What is the speed of a car moving in opposite direction if it meets the two cars at an interval of 4 min? a. \(60 \mathrm{~km} / \mathrm{h}\) b. \(15 \mathrm{~km} / \mathrm{h}\) c. \(30 \mathrm{~km} / \mathrm{h}\) d. \(45 \mathrm{~km} / \mathrm{h}\)

If two balls of same density but different masses are dropped from a height of \(100 \mathrm{~m}\), then (neglect air resistance) a. both will come together on the earth b. both will come late on the earth c. first will come first and second after that d. second will come first and first after that

A particle is dropped from rest from a large height. Assume \(g\) to be constant throughout the motion. The time taken by it to fall through successive distances of \(1 \mathrm{~m}\) each will be a. all equal, being equal to \(\sqrt{2 / g}\) second b. in the ratio of the square roots of the integers \(1,2,3, \ldots\) c. in the ratio of the difference in the square roots of the integers, i.e., \(\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}), \ldots\) d. in the ratio of the reciprocals of the square roots of the integers, i.e., \(\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \ldots\)

Two trains each travelling with a speed of \(37.5 \mathrm{~km} / \mathrm{h}\) are approaching each other on the same straight track. A bird that can fly at \(60 \mathrm{~km} / \mathrm{h}\) flies off from one train when they are \(90 \mathrm{~km}\) apart and heads directly for the other train. On reaching the other train it flies back to the first and so on. Total distance covered by the bird is a. \(90 \mathrm{~km}\) b. \(54 \mathrm{~km}\) c. \(36 \mathrm{~km}\) d. \(72 \mathrm{~km}\)

An elevator in which a man is standing is moving upwards with a speed of \(10 \mathrm{~m} / \mathrm{s}\). If the man drops a coin from a height of \(2.45 \mathrm{~m}\) from the floor of elevator, it reaches the floor of the elevator after a time \(\left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) a. \(\sqrt{2} s\) b. \(1 / \sqrt{2} \mathrm{~s}\) c. \(2 \mathrm{~s}\) d. \(1 / 2 \mathrm{~s}\)

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