Chapter 2: Problem 3
You check your odometer at the beginning of a day's driving and again at the end. Under what conditions would the difference between the two readings represent your displacement?
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Chapter 2: Problem 3
You check your odometer at the beginning of a day's driving and again at the end. Under what conditions would the difference between the two readings represent your displacement?
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You're speeding at \(85 \mathrm{km} / \mathrm{h}\) when you notice that you're only 10 m behind the car in front of you, which is moving at the legal speed limit of \(60 \mathrm{km} / \mathrm{h}\). You slam on your brakes, and your car negatively accelerates at \(4.2 \mathrm{m} / \mathrm{s}^{2}\). Assuming the other car continues at constant speed, will you collide? If so, at what relative speed? If not, what will be the distance between the cars at their closest approach?
If you know the initial velocity \(v_{0}\) and the initial and final heights \(y_{0}\) and \(y,\) you can use Equation 2.10 to solve for the time \(t\) when the object will be at height \(y .\) But the equation is quadratic in \(t,\) so you'll get two answers. Physically, why is this?
Derive Equation 2.10 by integrating Equation 2.7 over time. You'll have to interpret the constant of integration.
ThrustSSC, the world's first supersonic car, accelerates from rest to \(1000 \mathrm{km} / \mathrm{h}\) in \(16 \mathrm{s}\). What's its acceleration?
An object's position is given by \(x=b t+c t^{3},\) where \(b=1.50 \mathrm{m} / \mathrm{s}, c=0.640 \mathrm{m} / \mathrm{s}^{3},\) and \(t\) is time in seconds. To study the limiting process leading to the instantaneous velocity, calculate the object's average velocity over time intervals from (a) \(1.00 \mathrm{s}\) to \(3.00 \mathrm{s},\) (b) \(1.50 \mathrm{s}\) to \(2.50 \mathrm{s},\) and \((\mathrm{c}) 1.95 \mathrm{s}\) to \(2.05 \mathrm{s}.\) (d) Find the instantaneous velocity as a function of time by differentiating, and compare its value at 2 s with your average velocities.
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