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Passing. The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 \(\mathrm{m} / \mathrm{s}\) (about 45 \(\mathrm{mi} / \mathrm{h} )\) . Initially, the car is also traveling at 20.0 \(\mathrm{m} / \mathrm{s}\) and its front bumper is 24.0 \(\mathrm{m}\) behind the truck's rear bumper. The car accelerates at a constant 0.600 \(\mathrm{m} / \mathrm{s}^{2}\) , then pulls back into the truck's lane when the rear of the car is 26.0 \(\mathrm{m}\) ahead of the front of the truck. The car is 4.5 \(\mathrm{m}\) long and the truck is 21.0 \(\mathrm{m}\) long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?

Short Answer

Expert verified
Time: 3.57 s; Distance: 75.15 m; Final Speed: 22.14 m/s.

Step by step solution

01

Understand the Problem

We need to calculate the time it takes for the car to pass the truck, the distance the car travels during this time, and its final speed. Initially, the car's position is 24 m behind the truck and the relevant section in front of the truck (by which the car needs to overtake the truck) is 26 m.
02

Calculate the Total Overtaking Distance

The car needs to cover the initial gap of 24 m to pull up beside the truck, its own length of 4.5 m, the length of the truck (21.0 m), and extra 26 m beyond the front of the truck. Thus, the total distance to overtake is:\[ d_\text{total} = 24 + 4.5 + 21 + 26 = 75.5 \text{ m} \]
03

Set Up the Kinematic Equation

The car accelerates from an initial speed of 20.0 m/s with an acceleration of 0.600 m/s² to cover 75.5 m:\[ d = v_i t + \frac{1}{2} a t^2 \]Where:- \( v_i = 20.0 \text{ m/s} \)- \( a = 0.600 \text{ m/s}^2 \)- \( d = 75.5 \text{ m} \)
04

Solve for Time (t)

Plug in the known values into the kinematic equation:\[ 75.5 = 20.0 t + \frac{1}{2} \times 0.600 \times t^2 \]Simplifying:\[ 75.5 = 20.0t + 0.3t^2 \]This is a quadratic equation in the form of \( 0.3t^2 + 20.0t - 75.5 = 0 \). Solve using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 0.3 \), \( b = 20.0 \), and \( c = -75.5 \).
05

Calculate the Time (t) Using the Quadratic Formula

Using the quadratic formula:\[ t = \frac{-20.0 \pm \sqrt{20.0^2 - 4 \times 0.3 \times (-75.5)}}{2 \times 0.3} \]\[ t = \frac{-20.0 \pm \sqrt{400 + 90.6}}{0.6} \]\[ t = \frac{-20.0 \pm \sqrt{490.6}}{0.6} \]Calculate \( t \):\[ t \approx \frac{-20.0 + 22.14}{0.6} \approx 3.567 \text{ s} \] (we discard the negative root as time cannot be negative).
06

Calculate the Final Speed of the Car

The final speed \( v_f \) can be calculated using \( v_f = v_i + at \):\[ v_f = 20.0 + 0.600 \times 3.567 \approx 22.14 \text{ m/s} \].
07

Calculate the Distance Traveled by the Car

Using the formula for distance, \( d = v_i t + \frac{1}{2} a t^2 \), we have:\[ d = 20.0 \times 3.567 + \frac{1}{2} \times 0.600 \times (3.567)^2 \]Calculate:\[ d \approx 71.34 + 3.81 \approx 75.15 \text{ m} \] (rounded to match the conditions calculated previously).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Acceleration
Uniform acceleration occurs when an object's velocity changes at a constant rate. This means that the acceleration remains the same throughout the motion. In our exercise, the car's acceleration was given as 0.600 m/s², meaning every second, its speed increases by 0.600 meters per second.

This concept is crucial because it allows us to use kinematic equations, which are only applicable under conditions of constant acceleration. These equations help us find important quantities such as time, distance, and final velocity, which can be essential for solving problems in mechanics.
  • When acceleration is uniform, you can predict future motion using simple formulas.
  • Helps in determining how long it takes to reach a certain speed.
  • Calculating the distance covered during acceleration becomes easier.
Understanding uniform acceleration forms the foundation for solving many physics problems involving motion on linear paths, as shown by the car overtaking the truck.
Relative Motion
Relative motion is the calculation of the motion of an object with respect to another moving object. In our exercise, the car needed to pass a truck that was also in motion. Therefore, to determine the car's required overtaking path, we considered its movement relative to the truck.

Initially, both the truck and the car were moving at the same speed of 20.0 m/s. However, once the car started accelerating, its motion relative to the truck changed.
  • This involves adjusting calculations for both speed and distance in relation to another moving object.
  • Calculates the distance the car must travel to meet its target position ahead of the truck.
  • Often requires using additional terms to account for movements of all the objects involved.
This concept is useful for understanding how and when one moving object can overtake another, providing a deeper context into real-world driving scenarios.
Kinematic Equations
Kinematic equations describe the motion of objects and are used to derive unknown variables such as time, distance, or final velocity. These equations assume constant acceleration, which we had in the exercise.

The main kinematic equation used in this scenario was:\[ d = v_i t + \frac{1}{2} a t^2 \]where:
  • \( v_i \) is the initial velocity
  • \( a \) is the constant acceleration
  • \( t \) is the time taken
  • \( d \) is the distance covered

This equation was particularly useful for finding the time taken for the car to overtake the truck and calculating the distance covered during this process. By applying these formulas, kinematic problems become manageable, aiding students in predicting the outcomes of various motion scenarios.
Quadratic Equation
Solving a quadratic equation was essential in determining the time required for the car to pass the truck. Quadratic equations form when variables are squared, and they often appear in physics problems involving motion like our passing exercise.

In our case, the kinematic equation turned into a quadratic equation:\[ 0.3t^2 + 20.0t - 75.5 = 0 \]The quadratic formula \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]was used to solve the equation, with:
  • \( a = 0.3 \)
  • \( b = 20.0 \)
  • \( c = -75.5 \)
Using the quadratic formula helped us find the time (\( t \)) accurately, offering practical solutions when straightforward algebra isn't sufficient. Mastering how to work with quadratic equations is pivotal for anyone tackling complex physics problems, especially those exploring dynamics and kinematics.

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Most popular questions from this chapter

Cliff Height. You are climbing in the High Sierra where you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top and 10.0 \(\mathrm{s}\) later hear the sound of it hitting the ground at the foot of the cliff. (a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 \(\mathrm{m} / \mathrm{s} ?\) (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

A lunar lander is descending toward the moon's surface. Until the lander reaches the surface, its height above the surface of the moon is given by \(y(t)=b-c t+d t^{2},\) where \(b=800 \mathrm{m}\) is the initial height of the lander above the surface, \(c=60.0 \mathrm{m} / \mathrm{s},\) and \(d=1.05 \mathrm{m} / \mathrm{s}^{2} .\) (a) What is the initial velocity of the lander, at \(t=0 ?\) (b) What is the velocity of the lander just before it reaches the lunar surface?

Juggling Act. A juggler performs in a room whose ceiling is 3.0 \(\mathrm{m}\) above the level of his hands. He throws a ball upward so that it just reaches the ceiling. (a) What is the initial velocity of the ball? (b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, the juggler throws a second ball upward with two- thirds the initial velocity of the first. (c) How long after the second ball is thrown do the two balls pass each other? (d) At what distance above the juggler's hand do they pass each other?

Collision. The engineer of a passenger train traveling at 25.0 \(\mathrm{m} / \mathrm{s}\) sights a freight train whose caboose is 200 \(\mathrm{m}\) ahead onthe same track (Fig. P2.70). The freight train is traveling at 15.0 \(\mathrm{m} / \mathrm{s}\) in the same direction as the passenger train. The engineer of the passenger train immediately applies the brakes, causing a constant acceleration of 0.100 \(\mathrm{m} / \mathrm{s}^{2}\) in a direction opposite to the train's velocity, while the freight train continues with constant speed. Take \(x=0\) at the location of the front of the passenger train when the engineer applies the brakes. (a) Will the cows nearby witness a collision? (b) If so, where will it take place? (c) On a single graph, sketch the positions of the front of the passenger train and the back of the freight train.

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{m}+\) \(\left(4.80 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}-\left(0.100 \mathrm{m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x\) -t, \(v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{s}.\)

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