/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 A car and a motorcycle start fro... [FREE SOLUTION] | 91Ó°ÊÓ

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A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is \(25.0 \mathrm{~m}\) behind the car (v Fig. 2.27). The car accelerates at a uniform rate of \(3.70 \mathrm{~m} / \mathrm{s}^{2}\) and the motorcycle at a uniform rate of \(4.40 \mathrm{~m} / \mathrm{s}^{2}\) (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time? (c) How far ahead of the car will the motorcycle be \(2.00 \mathrm{~s}\) later? (Both vehicles are still accelerating.)

Short Answer

Expert verified
(a) 8.45 seconds, (b) Car: 132.16 m; Motorcycle: 157.16 m, (c) 38.60 m

Step by step solution

01

Understand the problem

We have two vehicles, a car and a motorcycle, starting from rest on a straight track. The motorcycle is initially 25.0 meters behind the car. Each vehicle accelerates uniformly at different rates: the car at \(3.70 \mathrm{~m/s^2}\) and the motorcycle at \(4.40 \mathrm{~m/s^2}\). Our task is to find the time it takes for the motorcycle to overtake the car, the distance they travel in that time, and the lead the motorcycle has \(2.00\) seconds after overtaking the car.
02

Set Equations of Motion

For both vehicles, we use the second equation of motion: \(s = ut + \frac{1}{2}at^2\). Since both start from rest, \(u = 0\) for both. Hence, for the car \(s_c = \frac{1}{2} \times 3.70 \times t^2\) and for the motorcycle \(s_m = \frac{1}{2} \times 4.40 \times t^2\). The motorcycle starts 25 m behind, so \(s_m = s_c + 25\).
03

Solve the Equation for Time

Equating the distances and setting \(s_m = s_c + 25\):\[\frac{1}{2} \times 4.40 \times t^2 = \frac{1}{2} \times 3.70 \times t^2 + 25\]Simplify to:\[2.20t^2 = 1.85t^2 + 25\]\[0.35t^2 = 25\]Solve for \(t\):\[t^2 = \frac{25}{0.35}\approx 71.43\]\[t \approx \sqrt{71.43}\approx 8.45 \text{ seconds}\].
04

Calculate Distance Traveled

Use the time found in the previous step to find the distance each vehicle travels by using the equation of motion.For the car:\[s_c = \frac{1}{2} \times 3.70 \times (8.45)^2 \approx 132.16 \text{ m}\]For the motorcycle:\[s_m = \frac{1}{2} \times 4.40 \times (8.45)^2 \approx 157.16 \text{ m}\]
05

Calculate Lead After 2 Seconds

The additional time is 2 seconds after the overtaking. The new time for each vehicle is \(t' = 8.45 + 2 = 10.45 \text{ s}\).Calculate new distances:For the car:\[s_c' = \frac{1}{2} \times 3.70 \times (10.45)^2 \approx 202.11 \text{ m}\].For the motorcycle:\[s_m' = \frac{1}{2} \times 4.40 \times (10.45)^2 \approx 240.71 \text{ m}\].The motorcycle's lead after 2 seconds is:\[\Delta s = s_m' - s_c' = 240.71 - 202.11 = 38.60 \text{ m}\].

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Key Concepts

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

Uniform Acceleration
In the world of kinematics, when an object moves along a straight line under constant acceleration, it is said to be undergoing uniform acceleration. This concept is crucial when analyzing the motion of vehicles like cars and motorcycles, as in our exercise here.

When acceleration is uniform, the object's velocity changes at a constant rate. This is due to external forces applied evenly over time, often seen in vehicles that consistently accelerate for increased speed. The motion can be described using the equations of motion, a set of formulas that relate various factors like time, velocity, and acceleration in the object's motion.

  • Uniform acceleration implies that the change in velocity is constant over equal time intervals.
  • In our scenario, the car accelerates at 3.70 m/s², while the motorcycle accelerates at a faster rate of 4.40 m/s², demonstrating uniformity.
  • Both vehicles start from rest, meaning their initial velocity is 0 m/s.
Understanding uniform acceleration allows us to predict how long it takes for one vehicle to surpass another when they start at different positions.
Equations of Motion
Equations of motion play a pivotal role in calculating how objects move through space and time, especially under the influence of uniform acceleration. These equations provide the tools needed to solve complex problems involving motion, like figuring out when and where vehicles will meet or travel a certain distance.

In kinematics, the basic equations of motion include parameters such as initial velocity, final velocity, acceleration, time, and displacement. The second equation of motion is particularly useful when objects start from rest and is given by:

\[s = ut + \frac{1}{2} at^2\]

In our example:

  • The car's displacement is expressed as \( s_c = \frac{1}{2} \times 3.70 \times t^2 \), assuming it starts from rest.
  • The motorcycle, also starting from rest, covers \( s_m = \frac{1}{2} \times 4.40 \times t^2 \).
  • The motorcycle initially lags behind by 25 meters, thus modifying its displacement equation to \( s_m = s_c + 25 \).
Solving this equation helps find the time both vehicles travel before the motorcycle overtakes the car.
Relative Motion
Relative motion is a fundamental concept in physics that describes how the position or velocity of one object is related to another. It's crucial when comparing the motions of two objects moving along the same or different paths.

In our exercise, relative motion becomes evident as we analyze the position and speed of both the car and motorcycle. Although they both start moving at the same time, the motorcycle must account for the initial gap of 25 meters that separates it from the car.

Why is Relative Motion Important?

  • It shows how the motorcycle, despite having a greater acceleration, needs time to catch up with the car.
  • The difference in acceleration allows calculation of the exact moment the motorcycle overtakes the car, which is after roughly 8.45 seconds.
  • Relative motion also helps determine how much further the motorcycle will be ahead after any given time, such as 2 seconds after overtaking.
Understanding relative motion allows students to gauge how objects change their positions relative to one another over time. It's a key concept to mastering problems involving two or more moving objects.

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Most popular questions from this chapter

Many highways with steep downhill areas have "runaway truck" inclined paths just off the main roadbed. These paths are designed so that if a vehicle's braking system gives out, the driver can steer it onto this incline (usually composed of loose gravel or sand). The idea is that the vehicle can then roll up the incline and come permanently and safely to rest with no need of a braking system. In one region of Hawaii the incline distance is \(300 \mathrm{~m}\) and provides a (constant) deceleration of \(2.50 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the maximum speed that a runaway vehicle can have as it enters the incline? (b) How long would such a vehicle take to come to rest? (c) Suppose another vehicle moving \(10 \mathrm{mi} / \mathrm{h}(4.47 \mathrm{~m} / \mathrm{s})\) faster than the maximum value enters the incline. What speed will it have as it leaves the gravel-filled area?

A motorboat traveling on a straight course slows uniformly from \(60 \mathrm{~km} / \mathrm{h}\) to \(40 \mathrm{~km} / \mathrm{h}\) in a distance of \(50 \mathrm{~m}\) What is the boat's acceleration?

A rifle bullet with a muzzle speed of \(330 \mathrm{~m} / \mathrm{s}\) is fired directly into a special dense material that stops the bullet in \(25.0 \mathrm{~cm}\). Assuming the bullet's deceleration to be constant, what is its magnitude?

A train makes a round trip on a straight, level track. The first half of the trip is \(300 \mathrm{~km}\) and is traveled at a speed of \(75 \mathrm{~km} / \mathrm{h}\). After a \(0.50 \mathrm{~h}\) layover, the train returns the \(300 \mathrm{~km}\) at a speed of \(85 \mathrm{~km} / \mathrm{h}\). What is the train's (a) average speed and (b) average velocity?

Two joggers run at the same average speed. Jogger A cuts directly north across the diameter of the circular track, while jogger B takes the full semicircle to meet his partner on the opposite side of the track. Assume their common average speed is \(2.70 \mathrm{~m} / \mathrm{s}\) and the track has a diameter of \(150 \mathrm{~m}\). (a) How many seconds ahead of jogger B does jogger A arrive? (b) How do their travel distances compare? (c) How do their displacements compare? (d) How do their average velocities compare?

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