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Chapter 2: Problem 246

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is $2 \mathrm{~V}$ (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

Short Answer

Expert verified
The acceleration when the velocity is 2V is \(20 \, \mathrm{m/s}^2\), which is option (A).

Step by step solution

01

Find the acceleration formula from the velocity formula

Differentiate the given velocity formula V = Ax + B with respect to time to find the acceleration formula. Recall that acceleration is the time derivative of velocity, so we are looking for a'(t), where a(t) represents acceleration, and V(t) represents velocity. a(t) = \(\frac{dV(t)}{dt} = \frac{d(Ax + B)}{dt}\)
02

Differentiate the velocity function

Differentiate the given velocity function with respect to time: \(\frac{dV}{dt} = \frac{d(Ax)}{dt} + \frac{d(B)}{dt}\) Since A and B are constants, the derivative of B with respect to time is 0. Also, to find the derivative of Ax with respect to time, we can use the chain rule: \(\frac{d(Ax)}{dt} = A\frac{dx}{dt}\) The derivative of x with respect to time is the velocity, so \(\frac{dx}{dt} = V\). Therefore, the acceleration function is: a(t) = A * V
03

Find the acceleration at V

We are given that the acceleration is 10 m/s² when the velocity is V. Use this information to determine the value of A in the acceleration function: 10 m/s² = A * V To find A, divide both sides by V: A = \(\frac{10 \, \mathrm{m/s}^2}{V}\)
04

Calculate the acceleration when the velocity is 2V

Now that we've found the value of A in terms of V, we'll calculate the acceleration when the velocity is 2V. Simply replace V in the acceleration function with 2V: a'(t) = A * (2V) a'(t) = \(\frac{10 \, \mathrm{m/s}^2}{V}\) * (2V) The V's cancel out, leaving: a'(t) = 20 m/s²
05

Determine the answer from the given choices:

The acceleration when the velocity is 2V is 20 m/s², which corresponds to option (A). Hence, the correct answer is (A) \(20 \, \mathrm{m/s}^2\).

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Most popular questions from this chapter

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is T Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Average velocity of the particle is (A) \((3 \mathrm{R} / \mathrm{T})\) (B) \((6 \mathrm{R} / \mathrm{T})\) (C) \((2 \mathrm{R} / \mathrm{T})\) (D) \((4 \mathrm{R} / \mathrm{T})\)

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) \((\mathrm{R} / \mathrm{T})\)

A particle has initial velocity \((2 \hat{1}+3 \hat{j}) \mathrm{ms}^{-1}\) and has acceleration \((\hat{1}+\hat{j}) \mathrm{ms}^{-2}\). Find the velocity of the particle after 2 second. (A) \((3 \hat{1}+5 \hat{j}) \mathrm{ms}^{-1}\) (B) \((4 \hat{i}+5 \hat{\jmath}) \mathrm{ms}^{-1}\) (C) \((3 \hat{1}+2 \hat{j}) \mathrm{ms}^{-1}\) (D) \((5 \hat{1}+4 \hat{j}) \mathrm{ms}^{-1}\)

A cartasian equation of a projectile is given by $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ Calculate its initial velocity. (A) \(\sqrt{10 \mathrm{~ms}^{-1}}\) (B) \(\sqrt{5 \mathrm{~ms}^{-1}}\) (C) \(\sqrt{2} \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

Velocity of particle \(\mathrm{A}\) with respect to particle \(\mathrm{B}\) is \(4(\mathrm{~m} / \mathrm{s})\) while they are moving in same direction. And it is \(10(\mathrm{~m} / \mathrm{s})\) while they are in opposite direction. What are the velocities of the particles with respect to the stationary frame of reference. (A) \(7 \mathrm{~ms}^{-1}, 3 \mathrm{~ms}^{-1}\) (B) \(4 \mathrm{~ms}^{-1}, 5 \mathrm{~ms}^{-1}\) (C) \(7 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\) (D) \(10 \mathrm{~ms}^{-1}, 4 \mathrm{~ms}^{-1}\)
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