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Chapter 2: Problem 309

A particle is moving in a xy plane with \(\mathrm{y}=2 \mathrm{x}\) and \(\mathrm{Vx}=2-\mathrm{t}\). Find \(\mathrm{Vy}\) at time \(\mathrm{t}=3\) second. (A) \(2 \mathrm{~ms}^{-1}\) (B) \(-3 \mathrm{~ms}^{-1}\) (C) \(+3 \mathrm{~ms}^{-1}\) (D) \(-2 \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The y-component of velocity (Vy) at time t = 3 seconds is (D) \( -2 \, ms^{-1} \).

Step by step solution

01

Differentiate y with respect to t

We have the trajectory equation y = 2x. We need to differentiate with respect to t to obtain Vy: \( \frac{dy}{dt} = Vy \) Applying the chain rule, we get: \( Vy = \frac{dy}{dx} * \frac{dx}{dt} \) Since \(\frac{dy}{dx}= 2\) (from y = 2x), the expression becomes: \( Vy = 2 * \frac{dx}{dt} \) But, \(\frac{dx}{dt} = Vx\), so: \( Vy = 2 * Vx \)
02

Substitute Vx into the expression

Now, we subsitute the expression for Vx, which is: \( Vx = 2 - t \) Into the expression for Vy, we get: \( Vy = 2 * (2 - t) \)
03

Calculate Vy at t = 3 seconds

Substitute t = 3 seconds into the Vy expression: \( Vy = 2 * (2 - 3) \) \( Vy = 2 * (-1) \) \( Vy = -2 \, ms^{-1} \) Hence the correct answer is: (D) \( -2 \, ms^{-1} \)

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Most popular questions from this chapter

Angle of projection, maximum height and time to reach the maximum height of a particle are \(\theta, \mathrm{H}\) and \(\mathrm{tm}\) respectively. Find the true relation. (A) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(\mathrm{H} / 2 \mathrm{~g})}\) (B) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(2 \mathrm{H} / \mathrm{g})}\) (C) \(\mathrm{t}_{\mathrm{m}}=\sqrt{(4 \mathrm{H} / \mathrm{g})}\) (D) \(t_{\mathrm{m}}=\sqrt{(\mathrm{H} / 4 \mathrm{~g})}\)

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