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Chapter 2: Problem 105

A motorcycle is moving at \(30 \mathrm{~m} / \mathrm{s}\) when the rider applies the brakes, giving the motorcycle a constant deceleration. During the \(3.0 \mathrm{~s}\) interval immediately after braking begins, the speed decreases to \(15 \mathrm{~m} / \mathrm{s}\). What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?

Short Answer

Expert verified
The motorcycle travels a total distance of 90 meters before stopping.

Step by step solution

01

Identify Known Values

We have the initial speed of the motorcycle \( u = 30 \, \text{m/s} \), the speed after 3 seconds \( v_1 = 15 \, \text{m/s} \), and the time interval \( t_1 = 3 \, \text{s} \). We need to find the total distance until the motorcycle stops.
02

Calculate Deceleration

Using the formula for acceleration \( a = \frac{v - u}{t} \), we find the deceleration between the initial speed and the speed after 3 seconds. Here, \( a = \frac{15 - 30}{3} = -5 \, \text{m/s}^2 \).
03

Distance Traveled in First 3 Seconds

Using the equation \( s_1 = ut + \frac{1}{2} a t^2 \), we calculate the distance traveled during the first 3 seconds. Thus, \( s_1 = 30 \times 3 + \frac{1}{2}(-5)(3^2) = 90 - 22.5 = 67.5 \, \text{m} \).
04

Find Time to Stop After First 3 Seconds

Using \( v = u + at \), we find time \( t_2 \) to stop. Initial speed after 3 seconds is \( 15 \, \text{m/s} \), final speed \( 0 \, \text{m/s} \), thus \( 0 = 15 + (-5)t_2 \), giving \( t_2 = 3 \, \text{s} \).
05

Distance Traveled Until Stop

Using \( s_2 = ut + \frac{1}{2} a t^2 \), calculate distance until stop from 15 m/s. Thus, \( s_2 = 15 \times 3 + \frac{1}{2}(-5)(3^2) = 45 - 22.5 = 22.5 \, \text{m} \).
06

Total Distance Traveled

Add the distances from the two intervals: \( s = s_1 + s_2 = 67.5 + 22.5 = 90 \, \text{m} \).

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Key Concepts

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

Deceleration
When a motorcycle begins braking, its speed doesn't just suddenly drop to zero. Instead, it reduces its speed at a uniform rate. This reduction in speed is known as deceleration, which is essentially negative acceleration.
To calculate deceleration, use the equation of motion: \[ a = \frac{v - u}{t} \]where:
  • \( a \) represents the acceleration (or deceleration, if negative),
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity, and
  • \( t \) is the time interval over which this change occurs.
In the context of the motorcycle, this equation helps determine how quickly it slows down from 30 m/s to 15 m/s within 3 seconds, resulting in a deceleration of -5 m/s虏. The negative sign confirms that the motorcycle is slowing down, not speeding up.
Distance traveled
The distance traveled during a deceleration period can be calculated using the equations of motion. The first part of the journey with a known duration and speed change is easy to calculate. Use the formula:\[ s = ut + \frac{1}{2} a t^2 \]Here:
  • \( s \) is the distance traveled,
  • \( u \) is the initial velocity (30 m/s),
  • \( a \) is the constant deceleration (-5 m/s虏),
  • \( t \) is the time interval (3 seconds).
For the motorcycle, this formula calculates that it travels 67.5 meters in the first 3 seconds. To find out how much additional distance the motorcycle travels until it stops, the same formula is used, adjusted for the new conditions after 3 seconds.
Equations of Motion
Understanding the equations of motion is crucial when calculating various kinematics problems including deceleration. These equations help us ascertain different variables related to motion, such as velocity, time, distance, and acceleration. The key three equations are:
  • \( v = u + at \): final velocity \( v \) after time \( t \),
  • \( s = ut + \frac{1}{2} a t^2 \): distance \( s \),
  • \( v^2 = u^2 + 2as \): helps calculate aspects without requiring time directly.
Using the right equation in appropriate scenarios allows us to solve complex problems in a structured manner. They provide a pathway from known variables to unknown outcomes, which is key in scenarios like the motorcycle slowing down and eventually stopping.

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Most popular questions from this chapter

In \(1 \mathrm{~km}\) races, runner 1 on track 1 (with time \(2 \mathrm{~min}, 27.95 \mathrm{~s}\) ) appears to be faster than runner 2 on track \(2(2 \mathrm{~min}, 28.15 \mathrm{~s})\). However, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large can \(L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of \(g\). (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T_{r}\) is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then \((\mathrm{d})\) what is \(T_{b}\) in terms of \(T_{r}\), and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing \(T_{r} .(\mathrm{f})\) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?

You are to drive \(300 \mathrm{~km}\) to an interview. The interview is at \(11: 15\) A.M. You plan to drive at \(100 \mathrm{~km} / \mathrm{h}\), so you leave at \(8: 00\) A.M. to allow some extra time. You drive at that speed for the first \(100 \mathrm{~km}\), but then construction work forces you to slow to \(40 \mathrm{~km} / \mathrm{h}\) for \(40 \mathrm{~km}\). What would be the least speed needed for the rest of the trip to arrive in time for the interview?

Suppose a rocket ship in deep space moves with constant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\), which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} ?(\mathrm{~b})\) How far will it travel in so doing?

You are arguing over a cell phone while trailing an unmarked police car by \(25 \mathrm{~m}\); both your car and the police car are traveling at \(110 \mathrm{~km} / \mathrm{h}\). Your argument diverts your attention from the police car for \(2.0 \mathrm{~s}\) (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that \(2.0 \mathrm{~s}\), the police officer begins braking suddenly at \(5.0 \mathrm{~m} / \mathrm{s}^{2} .(\mathrm{a})\) What is the separation between the two cars when your attention finally returns? Suppose that you take another \(0.40 \mathrm{~s}\) to realize your danger and begin braking. (b) If you too brake at \(5.0 \mathrm{~m} / \mathrm{s}^{2}\), what is your speed when you hit the police car?
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