/*! 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 22 A 745 i BMW car can brake to a s... [FREE SOLUTION] | 91Ó°ÊÓ

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A 745 i BMW car can brake to a stop in a distance of \(121 \mathrm{ft}\). from a speed of \(60.0 \mathrm{mi} / \mathrm{h} .\) To brake to a stop from a speed of \(80.0 \mathrm{mi} / \mathrm{h}\) requires a stopping distance of \(211 \mathrm{ft}\) What is the average braking acceleration for (a) \(60 \mathrm{mi} / \mathrm{h}\) to rest, (b) \(80 \mathrm{mi} / \mathrm{h}\) to rest, \((\mathrm{c}) 80 \mathrm{mi} / \mathrm{h}\) to \(60 \mathrm{mi} / \mathrm{h} ?\) Express the answers in \(\mathrm{mi} / \mathrm{h} / \mathrm{s}\) and in \(\mathrm{m} / \mathrm{s}^{2}\) .

Short Answer

Expert verified
The average braking acceleration is (a) -22 mi/hr/s and -9.8 m/s^2, (b) -24.6 mi/hr/s and -11 m/s^2, and (c) 6.7 mi/hr/s and 3 m/s^2 respectively.

Step by step solution

01

Convert speed from mi/hr to ft/s.

60 mi/hr = 60 * 5280 ft/3600 s = 88 ft/s; 80 mi/hr = 80 * 5280 ft/3600 s = 117.3 ft/s
02

Calculate average braking acceleration from 60 mi/hr to rest

For speed 60 mi/hr to 0, use the equation of motion \(v^2 = u^2 + 2as\). Rearrange it as \(a = (v^2 - u^2) / (2s)\). Substituting \(u = 88 ft/s\), \(v = 0\) and \(s = 121 ft\) gives \(a = -32.1 ft/s^2\).
03

Convert the acceleration to mi/hr/s

Acceleration \(a = -32.1 ft/s^2 = -32.1 * 3600 ft/5280 mi/hr/s = -22 mi/hr/s\)
04

Convert the acceleration to m/s^2

Acceleration \(a = -32.1 ft/s^2 = -32.1 * 0.3048 m/s^2 = -9.8 m/s^2\)
05

Repeat steps 2-4 for a speed of 80 mi/hr

Repeat steps 2-4 substituting \(u = 117.3 ft/s\) and \(s = 211 ft\). This gives the acceleration as -36.1 ft/s^2, -24.6 mi/hr/s and -11 m/s^2.
06

Find the average braking acceleration from 80 mi/hr to 60 mi/hr

Next, calculate from 80 to 60 mi/hr. Subtract the initial and final velocity to get a speed change of 29.3 ft/s. Then, use the equation \(a = Δv/Δt\). To find Δt, use the equation of motion \(s = ut + 1/2 at^2\). Then substitute \(s = 211-121 ft\), \(u = 88 ft/s\), and \(a = -32.1 ft/s^2\) to find Δt = 3.0 s. Finally, calculate \(a = Δv/Δt = 29.3 ft/s / 3.0 s = 9.8 ft/s^2\), which can be converted to 6.7 mi/hr/s and 3 m/s^2

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Key Concepts

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

Kinematic Equations
When analyzing motion, particularly in one dimension, we make use of kinematic equations. These equations describe the relationships between velocity, acceleration, time, and displacement. A foundational equation is \( v^2 = u^2 + 2as \), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.

These equations are paramount when solving problems regarding objects in motion, like a car braking to a stop. By rearranging the equation and inputting known values such as the stopping distance and initial speed, we can determine the car's average braking acceleration. It’s essential to ensure that all units are consistent to avoid errors in calculation.
Unit Conversion
Unit conversion is critical in physics problems because equations require consistency in units. Essentially, it's the process of converting measurements from one unit to another. In the problem at hand, speed is initially given in miles per hour (mi/hr) but for the kinematic equations to work, we need to convert that speed into feet per second (ft/s).

Such conversions require an understanding of the conversion factors, like knowing there are 5280 feet in a mile and 3600 seconds in an hour. After solving for acceleration in ft/s2, it’s often useful to convert to more standard units such as meters per second squared (m/s2) or miles per hour per second (mi/hr/s) for a clearer understanding of the car's deceleration rate.
Motion in One Dimension
Motion in one dimension refers to the movement of an object along a single straight path, where its position changes only along one specific axis—either the x or y axis in a coordinate system. In our problem, the BMW car's braking represents such motion.

We consider only the forward motion and the deceleration due to braking. The car's motion does not involve any turns or movement along a vertical path, simplifying the analysis to only considering one dimension – the horizontal path of the car.
Acceleration Calculation
Acceleration is defined as the rate of change of velocity of an object. To calculate the average braking acceleration, you use the kinematic equation rearranged to solve for acceleration: \(a = (v^2 - u^2) / (2s)\). Here, \(v\) is the final velocity (which is zero when the car stops), \(u\) is the initial velocity, and \(s\) is the distance over which the car stops.

This calculation yields acceleration in ft/s2, which can then be converted to other units of acceleration as demonstrated in the example. When the motion involves changing from one speed to another, such as from 80 mi/hr to 60 mi/hr, the acceleration can be calculated using the change in velocity (Δv) and the time taken (Δt) with the equation \(a = Δv/Δt\).

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Most popular questions from this chapter

A truck on a straight road starts from rest, accelerating at \(2.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) Then the truck travels for \(20.0 \mathrm{s}\) at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional \(5.00 \mathrm{s}\). (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

A car has an initial velocity \(v_{0}\) when the driver sees an obstacle in the road in front of him. His reaction time is \(\Delta t_{r}\) and the braking acceleration of the car is \(a\). Show that the total stopping distance is $$s_{\text {stop }}=v_{0} \Delta t_{r}-v_{0}^{2} / 2 a$$,Remember that \(a\) is a negative number.

The position of a particle moving along the \(x\) axis varies in time according to the expression \(x=3 t^{2},\) where \(x\) is in meters and \(t\) is in seconds. Evaluate its position (a) at \(t=3.00 \mathrm{s}\) and (b) at \(3.00 \mathrm{s}+\Delta t\). (c) Evaluate the limit of \(\Delta x / \Delta t\) as \(\Delta t\) approaches zero, to find the velocity at \(t=3.00 \mathrm{s}\).

A jet plane lands with a speed of \(100 \mathrm{m} / \mathrm{s}\) and can accelerate at a maximum rate of \(-5.00 \mathrm{m} / \mathrm{s}^{2}\) as it comes to rest.(a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this plane land on a small tropical island airport where the runway is \(0.800 \mathrm{km}\) long?

A hare and a tortoise compete in a race over a course 1.00 km long. The tortoise crawls straight and steadily at its maximum speed of \(0.200 \mathrm{m} / \mathrm{s}\) toward the finish line. The hare runs at its maximum speed of \(8.00 \mathrm{m} / \mathrm{s}\) toward the goal for \(0.800 \mathrm{km}\) and then stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise wins in a photo finish? Assume that, when moving, both animals move steadily at their respective maximum speeds.

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