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Chapter 2: Problem 244

The area under acceleration versus time graph for any time interval represents... (A) Initial velocity (B) final velocity (C) change in velocity in the time interval (D) Distance covered by the particle

Short Answer

Expert verified
The area under an acceleration versus time graph for any time interval represents the change in velocity in the time interval. This is because the area can be expressed as the product of the base (t) and the height (a), which can be rewritten as \(v - v0 = at\), illustrating the relationship between the area and the change in velocity.

Step by step solution

01

Recall basic equations of motion

Recall the following equation, which describes the relationship between acceleration (a), initial velocity (v0), final velocity (v), and time interval (t): \(v = v0 + at\)
02

Relationship between area and velocity

To understand the meaning of the area under the acceleration-time graph, let's consider the area of a basic shape, such as a rectangle. If the acceleration is constant throughout the time interval, the area of the rectangle is given by: Area = base × height = t × a Now, let's rewrite this in terms of velocity using the basic equation of motion: \(v - v0 = at\)
03

Interpret the area

By comparing the formula for the area under the acceleration-time graph with the equation of motion, we notice that the area represents the difference between the final velocity (v) and initial velocity (v0), which is the change in velocity during the time interval (t). Therefore, the correct answer is: (C) change in velocity in the time interval

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Most popular questions from this chapter

The resultant of two forces of magnitude \(2 \mathrm{~N}\) and \(3 \mathrm{~N}\) can never be. (A) \(4 \mathrm{~N}\) (B) \(1 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \((1 / 2) \mathrm{N}\)

Rohit completes a semicircular path of radius \(\mathrm{R}\) in 10 seconds. Calculate average speed and average velocity in \(\mathrm{ms}^{-1}\) (A) \([(2 \pi R) / 10],(2 R / 10)\) (B) \((\pi R / 10),(\mathrm{R} / 10)\) (C) \((\pi R / 10),(2 R / 10)\) (D) \([(2 \pi R) / 10],(\mathrm{R} / 10)\)

Find a unit vector from the followings. (A) \(\hat{\imath}+\hat{j}\) (B) \(\hat{\imath}-\hat{j}\) (C) \((1 / \sqrt{2}) \hat{\imath}+(1 / \sqrt{2}) \hat{\jmath}\) (D) \((1 / \sqrt{2}) \hat{1}-(1 / 2) \hat{\jmath}\)

Comprehensions type questions. A particle is moving in a circle of radius \(\mathrm{R}\) with constant speed. The time period of the particle is T Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Average speed of the particle is (A) \((\pi \mathrm{R} / 6 \mathrm{~T})\) (B) \([(2 \pi R) / 3 \mathrm{~T}]\) (C) \([(2 \pi R) / T]\) (D) \((\mathrm{R} / \mathrm{T})\)

\(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) are nonzero vectors. Which from the followings is true ? (A) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (B) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (C) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}\) (D) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}-\mathrm{B}^{2}\)
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