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A particle moving in a straight line covers half the distance with speed of \(3 \mathrm{~m} / \mathrm{s}\). The other half of the distance is covered in two equal time intervals and with speeds of \(4.5 \mathrm{~m} / \mathrm{s}\) and \(7.5 \mathrm{~m} / \mathrm{s}\) respectively. The average speed of the particle during this motion is (a) \(4.0 \mathrm{~m} / \mathrm{s}\) (b) \(5.0 \mathrm{~m} / \mathrm{s}\) (c) \(5.5 \mathrm{~m} / \mathrm{s}\) (d) \(4.8 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The average speed is \(4.0 \, \mathrm{m/s}\) (option a).

Step by step solution

01

Understanding the Problem

We have a particle moving in a straight line where the total distance is split into two halves. The first half is covered at a speed of \(3 \, \mathrm{m/s}\). The second half is divided into two equal time intervals with speeds of \(4.5 \, \mathrm{m/s}\) and \(7.5 \, \mathrm{m/s}\). We need to calculate the average speed of the particle for the entire journey.
02

Setting Up the Problem

Let's denote the total distance as \(2d\). Hence, one half of the distance, \(d\), is covered at \(3 \, \mathrm{m/s}\). The distance for the second half is also \(d\), split into two segments covered in equal time at \(4.5 \, \mathrm{m/s}\) and \(7.5 \, \mathrm{m/s}\).
03

Calculating Time for First Half

The time \(t_1\) to travel the first half of the distance \(d\) at \(3 \, \mathrm{m/s}\) is \(t_1 = \frac{d}{3}\).
04

Calculating Distance Covered in Each Part of the Second Half

For the second half of the distance \(d\): Suppose it is covered in two equal time intervals: \(t_2\) with a speed of \(4.5 \, \mathrm{m/s}\) and \(t_3\) with a speed of \(7.5 \, \mathrm{m/s}\). In equal time \(t/2\), the distances covered are \(d_1 = 4.5 \left( \frac{t}{2} \right)\) and \(d_2 = 7.5 \left( \frac{t}{2} \right)\).
05

Determining the Time for Each Bit of the Second Half

Given \(d_1 + d_2 = d\), we have:\[4.5 \left( \frac{t}{2} \right) + 7.5 \left( \frac{t}{2} \right) = d\]\[6t = d\]Thus, the time spent in each part of this half is \(t/2 = \frac{d}{12}\). Therefore, \(t = \frac{d}{6}\).
06

Calculating Total Time

Total time \(T\) is \(t_1 + t_2 + t_3 = \frac{d}{3} + \frac{d}{12} + \frac{d}{12}\). Simplifying this gives:\[T = \frac{d}{3} + \frac{d}{6}\]\[= \frac{2d}{6} + \frac{d}{6}\]\[= \frac{3d}{6}\]\[= \frac{d}{2}\]
07

Finding the Average Speed

Average speed \(v_{avg}\) is given by the formula:\[v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}\]The total distance is \(2d\), and total time is \(\frac{d}{2}\).\[v_{avg} = \frac{2d}{\frac{d}{2}}\]\[= 4 \, \mathrm{m/s}\].
08

Conclusion

The average speed of the particle during its motion is \(4.0 \, \mathrm{m/s}\), which corresponds to option (a).

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Key Concepts

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

Kinematics
When we talk about kinematics, we're diving into the branch of physics dealing with motion. Kinematics tells us how an object moves without considering the causes of this motion due to forces. It focuses purely on the geometry of motion. Think of it like a storyteller, telling us how fast and in what direction an object moves, and how long it takes to get from one point to another.
Kinematic analysis relies heavily on clearly defined parameters like speed, velocity, distance traveled, and time taken. These parameters give us a comprehensive picture of a particle's movement. For example, in the context of the problem above, kinematics helps explain how the speed varies across different portions of the journey and calculates the total average speed.
Uniform Motion
Uniform motion is a basic yet powerful concept in physics. It's when an object covers equal distances in equal time intervals, essentially moving at a constant speed. Imagine driving a car on a highway where the speedometer consistently reads the same speed. This is uniform motion.
In the exercise, we encounter multiple speeds, but the first half of the journey at 3 m/s can be seen as a scenario where uniform motion is applicable. Knowing how to identify and work with uniform motion helps in simplifying complex movement patterns, breaking them down into smaller segments that we can easily calculate and understand.
Time Intervals
Time intervals are the slices of time during which specific parts of motion occur. Understanding time intervals is crucial in breaking down a motion into manageable parts, each with its own speed and distance covered. This becomes handy in kinematic problems, like the one in our exercise, where different speeds are applied in different segments.
During the second half of the distance in the problem, the motion is split into two equal time intervals. Each is crucial to find out how long the particle travels at each particular speed. Calculating these intervals helps us determine the durations needed at various speeds to complete segments of the journey, eventually leading to the correct average speed calculation.
Speed Calculation
When solving motion problems, speed calculation acts as a tool to quantify how fast an object travels over a given period or distance. In everyday terms, it's just how quickly something moves from one place to another.
To calculate speed, we generally use the formula: \(v = \frac{d}{t}\), where \(v\) is speed, \(d\) is distance, and \(t\) is time. For an average speed calculation, as seen in the exercise, we use the entire journey's total distance and total time to find \(v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}}\). Understanding and applying this formula helps determine a realistic average speed, which is key in predicting travel times and planning efficient routes in both practical and application-based scenarios.

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Most popular questions from this chapter

From a balloon rising vertically upwards at \(5 \mathrm{~m} / \mathrm{s}\) a stone is thrown up at \(10 \mathrm{~m} / \mathrm{s}\) relative to the balloon. Its velocity with respect to ground after \(2 \mathrm{~s}\) is (assume \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(\underline{0}\) (b) \(20 \mathrm{~m} / \mathrm{s}\) (c) \(10 \mathrm{~m} / \mathrm{s}\) (d) \(5 \mathrm{~m} / \mathrm{s}\)

A ball is dropped on the floor from a height of \(10 \mathrm{~m}\). It rebounds to a height of \(2.5 \mathrm{~m}\). If the ball is in contact with the floor for \(0.01 \mathrm{~s}\), the average acceleration during contact is nearly (Take \(g=10 \mathrm{~ms}^{-2}\) ) \(\begin{array}{ll}\text { (a) } 500 \sqrt{2} \mathrm{~ms}^{-2} \text { upwards } & \text { (b) } 1800 \mathrm{~ms}^{-2} \text { downwards }\end{array}\) \(\begin{array}{ll}\text { (c) } 1500 \sqrt{5} \mathrm{~ms}^{-2} \text { upwards } & \text { (d) } 1500 \sqrt{2} \mathrm{~ms}^{-2} \text { downwards }\end{array}\)

An elevator ascends with an upward acceleration of \(2.0 \mathrm{~ms}^{-2}\). At the instant its upward speed is \(2.5 \mathrm{~ms}^{-1}\), loose bolt is dropped from the ceiling of the elevator \(3.0 \mathrm{~m}\) from the floor. If \(g=10 \mathrm{~ms}^{-2}\), then (a) the time of flight of the bolt from the ceiling to floor of the elevator is \(0.11 \mathrm{~s}\) (b) the displacement of the bolt during the free fall relative to the elevator shaft is \(0.75 \mathrm{~m}\) (c) the distance covered by the bolt during the free fall relative to the elevator shaft is \(1.38 \mathrm{~m}\) (d) the distance covered by the bolt during the free fall relative to the elevator shaft is \(2.52 \mathrm{~m}\)

Three particles start from the origin at the same time, one with a velocity \(v_{1}\) along \(x\)-axis, the second along the \(y\)-axis with a velocity \(v_{2}\) and the third along \(x=y\) line. The velocity of the third so that the three may always lie on the same line is (a) \(\frac{v_{1} v_{2}}{v_{1}+v_{2}}\) (b) \(\frac{\sqrt{2} v_{1} v_{2}}{v_{1}+v_{2}}\) (c) \(\frac{\sqrt{3} v_{1} v_{2}}{v_{1}+v_{2}}\) (d) zero

A jet airplane travelling at a speed of \(500 \mathrm{~km} / \mathrm{h}\) ejects its products of combustion at the speed of \(1500 \mathrm{~km} / \mathrm{h}\) relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? [NCERT] (a) \(-1000 \mathrm{~km} / \mathrm{h}\) (b) \(1000 \mathrm{~km} / \mathrm{h}\) (c) \(100 \mathrm{~km} / \mathrm{h}\) (d) \(-100 \mathrm{~km} / \mathrm{h}\)

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