/*! 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 29 The driver of a car slams on the... [FREE SOLUTION] | 91Ó°ÊÓ

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The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of \(-5.60 \mathrm{m} / \mathrm{s}^{2}\) for \(4.20 \mathrm{s}\), making straight skid marks \(62.4 \mathrm{m}\) long ending at the tree. With what speed does the car then strike the tree?

Short Answer

Expert verified
The car strikes the tree at 2.3 m/s.

Step by step solution

01

Identify the knowns and unknowns

The given acceleration \(a = -5.60 m/s^2\), displacement \(s = 62.4 m\), and time \(t = 4.20 s\). The velocity \(v\) is what we need to find. The negative sign indicates deceleration or slowing down. The initial velocity \(u\) is unknown.
02

Apply the uniform motion equation

We solve for \(u\) from the equation \(s = ut + 0.5at^2\). We rearrange the equation to solve for \(u\), giving \(u = (s - 0.5at^2) / t\). Substituting the given values, \(u = (62.4 m - 0.5(-5.60 m/s^2)(4.20 s)^2)/4.20 s = 26.05 m/s\). We find the initial velocity of the car to be 26.05 m/s.
03

Solve for the final velocity with which the car strikes the tree

We solve for \(v\) using the equation \(v = u + at\). We substitute the given values, giving \(v = 26.05 m/s + (-5.60 m/s^2)(4.20 s) = 2.3 m/s\). The velocity with which the car strikes the tree is 2.3 m/s

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Key Concepts

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

Uniform Acceleration
In kinematics, uniform acceleration refers to a constant rate of change in velocity per time, which can be either speeding up or slowing down of an object. In simple terms, if an object is under uniform acceleration, it means that its velocity is changing at a steady pace as time goes on. An acceleration is described by its magnitude and its direction – negative acceleration (also known as deceleration) indicates that the object is slowing down, while positive acceleration indicates it is speeding up.

For instance, in our example problem, the car exhibits a uniform deceleration (acceleration with a negative value) of \( -5.60 \text{ m/s}^2 \) as the driver applies the brakes. This 'constant' aspect is crucial when using equations of motion to predict the future position or velocity of an object. One of the core equations derived from the principles of uniform acceleration is \( s = ut + \frac{1}{2}at^2 \) where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Initial Velocity
Initial velocity can be thought of as the speed at which an object begins its journey before any acceleration or deceleration is applied. Knowing the initial velocity is vital as it sets the stage for calculating changes in motion that occur due to acceleration. In the context of our problem, the initial velocity (\( u \) ) represents the speed right before the driver begins to brake.

To calculate initial velocity when dealing with uniform acceleration, we can manipulate the equation \( s = ut + \frac{1}{2}at^2 \) to solve for \( u \). Rearranging the equation, we get \( u = \frac{s - \frac{1}{2}at^2}{t} \). By substituting the distance (\( s = 62.4\text{ m} \)), the acceleration (\( a = -5.60 \text{ m/s}^2 \) ), and the time (\( t = 4.20 \text{ s} \) ), we can find the car's initial velocity just as the driver started to brake.
Final Velocity
Final velocity refers to the speed of an object at the conclusion of a period of acceleration; it's where the object ends up after a change in speed has occurred. In kinematic problems, it is often what is being solved for after initial velocity and acceleration have been identified. The formula to compute the final velocity (\( v \) ) when an initial velocity (\( u \) ) and a constant acceleration (\( a \) ) are applied over a period of time (\( t \) ) is \( v = u + at \).

For the example at hand, after finding the initial velocity, you can find how fast the car was moving when it hit the tree (the final velocity) by plugging the initial velocity, acceleration, and time into the formula. Doing so reveals that the car's final velocity was \( 2.3 \text{ m/s} \), indicating a significant reduction in speed, thanks to the brakes being applied over the given timeframe. These calculations are essential for understanding the dynamics of motion and can be used in a variety of practical situations, including accident reconstructions and safety analyses.

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Most popular questions from this chapter

Draw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, \((c)\) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform; that is, if the speed were not changing at a constant rate?

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at \(71.5 \mathrm{m} / \mathrm{s} .\) The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of \(250 \mathrm{m}\). He spends \(5.00 \mathrm{s}\) in the pit and then accelerates out, reaching his previous speed of \(71.5 \mathrm{m} / \mathrm{s}\) after a distance of \(350 \mathrm{m} .\) At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the height it reaches.

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