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Chapter 3: Problem 45

A particle's constant acceleration is north at $100 \mathrm{m} / \mathrm{s}^{2}\( At \)t=0,\( its velocity vector is \)60 \mathrm{m} / \mathrm{s}$ east. At what time will the magnitude of the velocity be $100 \mathrm{m} / \mathrm{s} ?$

Short Answer

Expert verified
Answer: The magnitude of the velocity of the particle will be 100 m/s at 0.8 seconds.

Step by step solution

01

Break down the velocity vector into components

Given that the initial velocity is \(60 \mathrm{m}/\mathrm{s}\) east, the eastward component of velocity is \(60 \mathrm{m}/\mathrm{s}\). Since there is no initial northward velocity, the northward component of velocity at \(t=0\) is \(0 \mathrm{m}/\mathrm{s}\). Thus, the initial velocity vector can be written as: $$\vec{v}(0) = 60i \mathrm{m}/\mathrm{s}.$$
02

Find the expression for velocity in terms of time

Given that the acceleration is constant and in the north direction, it can be written as: $$\vec{a} = 100j \mathrm{m}/\mathrm{s}^2.$$ Now, using the definition of acceleration, $$\vec{a} = \frac{d\vec{v}}{dt}$$Integrating both sides, we can find the velocity as a function of time:$$\vec{v}(t) = \int \vec{a} dt = \int 100j dt = \vec{v}(0) + 100tj = 60i + 100tj.$$Here, we have used the initial condition \(\vec{v}(0) = 60i \mathrm{m}/\mathrm{s}\) to solve for the integration constant.
03

Calculate the magnitude of the velocity and find when it equals the given value

Find the magnitude of the velocity vector:$$|\vec{v}(t)| = \sqrt{(60)^2 + (100t)^2} = \sqrt{3600 + 10000t^2}.$$We are asked to find the time at which the magnitude of the velocity becomes \(100 \mathrm{m}/\mathrm{s}\):$$100 = \sqrt{3600 + 10000t^2}$$Square both sides to get rid of the square root:$$10000 = 3600 + 10000t^2$$Solve for \(t^2\):$$t^2 = \frac{10000-3600}{10000} = 0.64$$Take the square root to get the value of \(t\):$$t = \sqrt{0.64} = 0.8 \mathrm{s}$$ Thus, the magnitude of the velocity will be \(100 \mathrm{m}/\mathrm{s}\) at \(t=0.8\mathrm{s}\).

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Most popular questions from this chapter

A runner times his speed around a track with a circumference of $0.50 \mathrm{mi} .\( He finds that he has run a distance of \)1.00 \mathrm{mi}$ in 4.0 min. What is his (a) average speed and (b) average velocity magnitude in \(\mathrm{m} / \mathrm{s}\) ?
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