(i) When a body travels in straight line path, its distance and displacements are equal and hence in a given time, velocity is equal to speed. (ii) Average velocity of a body may be equal to zero if the total displacement of is body is equal to zero, but average speed will not be equal to zero unless and until distance covered is zero (i.e., body at rest). (iii) To describe the velocity, direction is necessary as velocity is speed in a specific direction.

Short Answer

Expert verified
Explain the relationship between speed and velocity in this case. Answer: Yes, the distance and displacement can be equal in straight line motion. In such motion, the length covered by the body (distance) equals the length between the initial and final positions (displacement). In this case, since distance and displacement are equal, the speed and velocity also become equal. Speed represents the rate at which distance is covered, while velocity represents the rate at which displacement occurs.

Step by step solution

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(i) Distance, Displacement, and Velocity in Straight Line Motion

When a body travels in a straight line, its distance and displacement are equal. Distance is the total length covered by the body, while displacement is the length between the initial and final positions. In straight line motion, the length covered equals the length between the initial and final positions. Since both distance and displacement are equal in straight line motion, the speed and velocity are also equal, as speed is the rate at which distance is covered, while velocity is the rate at which displacement occurs.
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(ii) Average Velocity Equals Zero and Average Speed Non-Zero

Average velocity can be equal to zero if the total displacement of the body is zero. This can happen if the object returns to its starting point, as the distance between initial and final positions will be zero. However, the average speed will not be zero unless the distance covered is zero which happens when the body is at rest. The average speed is the total distance covered divided by the time taken, so if the distance is non-zero, the average speed will also be non-zero.
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(iii) Why Direction is Necessary for Describing Velocity

Velocity is a vector quantity, which means it has both magnitude (speed) and direction. To fully describe velocity, the direction must be specified. In contrast, speed is a scalar quantity and only has magnitude, and thus direction is not necessary when describing speed. Including direction in the description of velocity helps us understand how an object is moving; without direction, we would only know the rate at which the object is moving but not where it is heading.

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Key Concepts

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

Distance and Displacement
Imagine you're on a hike, following a trail that turns and winds through the mountains. The distance you cover is like the total length of the trail you've hiked, including every twist and turn. On the other hand, displacement is a straight shot from your starting point to your endpoint, regardless of the path you took. In physics for class 7 students, we deal with these concepts to understand motion.

When motion occurs in a straight line, it's like taking a shortcut that doesn't turn or deviate; your distance and displacement are the same, as there's no difference between the path taken and the direct line from start to finish. This is crucial because it simplifies the calculations for speed and velocity, two concepts closely linked to distance and displacement.
Velocity and Speed
Speed is how fast you're going without worrying about your destination—think of it like the number you see on a speedometer in a car. Now, if you add your destination to the mix, saying, 'I’m heading north at this speed,' you're talking about velocity. It's speed with a direction. Both are important when studying physics motion concepts in class 7.

Comparing Speed and Velocity

Speed is a scalar quantity—it has just the magnitude, like '60 mph.' Velocity, however, is a vector quantity—it has both magnitude and direction, like '60 mph north.' Knowing the velocity is more informative because it tells us the speed an object is moving and in which direction.
Average Velocity and Average Speed
Let's say you're watching a one-hour movie at the cinema. If you get up and walk around but end up back in your seat by the end of the movie, your average velocity is zero because you haven't changed your position from start to finish. But your average speed is not zero because you indeed moved around—this is the total distance you walked divided by the movie's duration.

Average speed and velocity are like summaries of your entire trip. Average speed is straightforward—it's how fast you've been moving on average. Average velocity adds more context by considering your overall change in position. If you start and end in the same place, like returning to your seat, average velocity says, 'You haven't gone anywhere on average,' even though you moved around.
Vector and Scalar Quantities
Scalar quantities are the simple numbers in physics; they tell you 'how much' but not 'which way.' Examples include mass, speed, and distance. They're like the money in your wallet—you know the amount, but it doesn't tell you where you're going to spend it.

In contrast, vector quantities give you more information—they have both a magnitude and a direction, like an arrow. Think of it as giving you money and a shopping list; it tells you 'how much' and 'where to go.' This would include quantities such as velocity, displacement, and force. Understanding the difference between these two types of quantities is essential as they tell us different things about movement and can change the meaning of a physics problem entirely.

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Most popular questions from this chapter

Fill in the Blanks. \(\frac{1}{1000}\) 1000 millisecond \(=1 \mathrm{~s}\) \(\Rightarrow 1\) millisecond \(=\frac{1}{1000}\) th part of a second.

We know \(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \Rightarrow \mathrm{T} \Rightarrow \sqrt{\ell}\) \(\mathrm{T}^{2}=4 \pi^{2} \frac{\ell}{\mathrm{g}} \Rightarrow \mathrm{T}^{2} \alpha \ell\) From above relationships, it is clear that as the length of simple pendulum increases, time period also increases and vice versa.

Let the distance travelled by car with \(42 \mathrm{~km} \mathrm{~h}^{-1}\) for 20 minutes be \(\mathrm{S}\), and distance travelled with 60 \(\mathrm{km} \mathrm{h}^{-1}\) for 30 minutes be \(\mathrm{S}_{2}\) \(\Rightarrow \mathrm{S}_{1}=42 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{3} \mathrm{~h}=14 \mathrm{~km}\) \(\Rightarrow \mathrm{S}_{2}=60 \mathrm{~km} \mathrm{~h}^{-1} \times \frac{1}{2} \mathrm{~h}=30 \mathrm{~km}\) Total distance \(\mathrm{S}=\mathrm{S}_{1}+\mathrm{S}_{2}=14+30=44 \mathrm{~km}\)

We know, \(\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\) The time period of seconds pendulum, \(\mathrm{T}_{1}=2 \mathrm{~s}\) The length of the second pendulum \(\ell_{1}=100 \mathrm{~cm}=1 \mathrm{~m}\) Now, the new length of the pendulum \(\ell_{2}=2 \ell_{1}\) \(\Rightarrow \ell_{2}=200 \mathrm{~cm}=2 \mathrm{~m}\) Let the new time period of the pendulum be \(=\mathrm{T}_{2}\) \(\Rightarrow \mathrm{T} \alpha \sqrt{l}\) \(\Rightarrow \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{l_{1}}{l_{2}}} \Rightarrow \mathrm{T}_{2}=\mathrm{T}_{1} \sqrt{\frac{l_{2}}{l_{1}}}\) \(\mathrm{T}_{2}=2 \times \sqrt{\frac{2 l_{1}}{l_{1}}}=2 \sqrt{2} \mathrm{~s}\) The time period becomes \(2 \sqrt{2}\) times the original one.

(i) The unit of speed and velocity is \(\mathrm{m} \mathrm{s}^{-1}\). (ii) If a body is moving in a straight line path then its speed is equal to velocity \(50 \mathrm{~m} \mathrm{~s}^{-1}=50 \times \frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}=180 \mathrm{~km} \mathrm{~h}^{-1}\)

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