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Chapter 3: Problem 67

A car is driving directly north on the freeway at a speed of $110 \mathrm{km} / \mathrm{h}$ and a truck is leaving the freeway driving \(85 \mathrm{km} / \mathrm{h}\) in a direction that is \(35^{\circ}\) west of north. What is the velocity of the truck relative to the car?

Short Answer

Expert verified
Answer: The relative velocity of the truck with respect to the car is 63.31 km/h, 39.9° east of south.

Step by step solution

01

Find the x and y components of each velocity vector

We will begin by finding the x and y components of the velocities of the car and truck. For the car, the velocity vector is directly north, so we only have a y-component, which is \(110 \mathrm{km/h}\). The truck's velocity has both x and y components since it is traveling \(35^\circ\) west of north. Truck's x-component of velocity: \(-v_t\sin(35^\circ)\) Truck's y-component of velocity: \(v_t\cos(35^\circ)\) Where \(v_t = 85 \mathrm{km/h}\). Now, we will calculate the values of these components.
02

Calculate the x and y components

Using the given truck speed and direction, we can find the components of its velocity. x-component: \((-85 \sin(35^\circ)) \mathrm{km/h} = -48.65 \mathrm{km/h}\) y-component: \((85 \cos(35^\circ)) \mathrm{km/h} = 69.71 \mathrm{km/h}\) Now we have: Car velocity = \((0, 110) \mathrm{km/h}\) Truck velocity = \((-48.65, 69.71) \mathrm{km/h}\)
03

Calculate the relative velocity components

Now, to find the relative velocity of the truck with respect to the car, we will subtract the car's velocity components from the truck's velocity components. Relative x-component: \((-48.65 - 0) \mathrm{km/h} = -48.65 \mathrm{km/h}\) Relative y-component: \((69.71 - 110) \mathrm{km/h} = -40.29 \mathrm{km/h}\) So the relative velocity components are \((-48.65, -40.29) \mathrm{km/h}\).
04

Calculate the magnitude and direction of the relative velocity

Now, we will find the magnitude and direction of the relative velocity. Magnitude: \(\sqrt{(-48.65)^2 + (-40.29)^2} \mathrm{km/h} = 63.31 \mathrm{km/h}\) Direction: \(\tan^{-1}(\frac{-40.29}{-48.65}) = 39.9^\circ\) east of south Finally, we have the relative velocity of the truck with respect to the car is \(63.31 \mathrm{km/h}\), \(39.9^\circ\) east of south.

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Most popular questions from this chapter

A jetliner flies east for \(600.0 \mathrm{km},\) then turns \(30.0^{\circ}\) toward the south and flies another \(300.0 \mathrm{km} .\) (a) How far is the plane from its starting point? (b) In what direction could the jetliner have flown directly to the same destination (in a straight-line path)? (c) If the jetliner flew at a constant speed of \(400.0 \mathrm{km} / \mathrm{h}\), how long did the trip take? (d) Moving at the same speed, how long would the direct flight have taken?
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