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Chapter 2: Problem 146

A particle starts from the origin at \(t=0 \mathrm{~s}\) with a velocity of \(10.0 \hat{j} \mathrm{~m} / \mathrm{s}\) and moves in the \(x-y\) plane with a constant acceleration of \((8.0 \hat{i}+2.0 \hat{j}) \mathrm{ms}^{-2}\) At a time when \(x\)-coordinate of the particle \(16 \mathrm{~m}\), (A) \(24 \mathrm{~m}\) (B) \(22 \mathrm{~m}\) (C) \(20 \mathrm{~m}\) (D) None

Short Answer

Expert verified
The y-coordinate of the particle when the x-coordinate is 16 m is (A) \(24 \mathrm{~m}\).

Step by step solution

01

Determine the equations for x and y positions at any time t

We have the given initial velocity and acceleration vectors: \(v_0 = 10.0 \hat{j} \mathrm{~m} / \mathrm{s}\) \(a = (8.0\hat{i} + 2.0\hat{j})\mathrm{ms}^{-2}\) We can find the position vector (x, y) at any time t using the equation: \(r(t) = r_0 + v_0 t + \frac{1}{2} a t^2\) Since the particle starts from the origin, the initial position \(r_0\) is at the origin (0, 0). Therefore, we can rewrite the equation for x and y components as: \(x(t) = 0 + 0 + \frac{1}{2} (8.0)t^2 = 4t^2\) \(y(t) = 0 + 10t + \frac{1}{2} (2.0)t^2 = 10t + t^2\)
02

Solve for the time when x-coordinate is 16 m

We are given that at a certain time, the x-coordinate of the particle is 16 m. So, we can set \(x(t) = 16\) and solve for time t: \(16 = 4t^2\) \(t^2 = 4\) \(t = 2 \mathrm{s}\) The particle will have an x-coordinate of 16 m at t = 2 s.
03

Calculate the y-coordinate when x-coordinate is 16 m

Now we have the time when the x-coordinate is 16 m, we can find the y-coordinate using the equation for \(y(t)\): \(y(2) = 10(2) + (2)^2\) \(y(2) = 20 + 4\) \(y(2) = 24 \mathrm{m}\) This matches option (A), so the correct answer is: (A) \(24 \mathrm{~m}\)

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