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An automobile traveling \(35 \mathrm{mi} / \mathrm{h}(=56 \mathrm{~km} / \mathrm{h})\) is \(110 \mathrm{ft}(=34\) \(\mathrm{m}\) ) from a barrier when the driver slams on the brakes. Four seconds later the car hits the barrier. ( \(a\) ) What was the automobile's constant deceleration before impact? (b) How fast was the car traveling at impact?

Short Answer

Expert verified
(a) The constant deceleration of the automobile before impact was -3.89 \(m/s^2\). (b) The speed of the car at impact was 0 km/h.

Step by step solution

01

Identify given variables and convert to appropriate units

First, identify all given variables: initial velocity \(v_i = 35 \mathrm{mi} / \mathrm{h} = 56 \mathrm{km} / \mathrm{h} = 15.56 \mathrm{m/s}\), displacement \(d = 34 \mathrm{m}\), and time \(t = 4 \mathrm{s}\). For simplicity, convert the initial velocity from km/h to m/s by multiplying by \(0.278\).
02

Calculate deceleration using second equation of motion

The second equation of motion is \(d = v_it + \frac{1}{2}at^2\), where \(a\) is acceleration (or in this case, deceleration). To find \(a\), rearrange the equation to get \(a = \frac{2(d - v_it)}{t^2}\). Substitute the values of \(d, v_i,\) and \(t\) to calculate \(a\).
03

Calculate final velocity

The first equation of motion is \(v_f = v_i + at\), where \(v_f\) is the final velocity. Using the acceleration (deceleration) calculated from Step 2 and the given initial velocity and time, calculate the final velocity \(v_f\).
04

Conversion of final velocity to km/h

For consistency with the question statement, convert the final velocity \(v_f\) from m/s to km/h by dividing by \(0.278\).

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Key Concepts

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

Equations of Motion
The equations of motion are fundamental in studying how objects move in physics, particularly when describing an object's position, velocity, and acceleration over time under constant acceleration. In the context of deceleration problems, these equations allow us to connect these variables and accurately predict a vehicle's stopping distance and speed after applying brakes.

For example, the second equation of motion \( d = v_it + \frac{1}{2}at^2 \) incorporates initial velocity (\( v_i \)), acceleration (\( a \)), time (\( t \)), and displacement (\( d \)) to form a formula that can be rearranged to solve for various unknowns. When a car decelerates, the same equations apply, but the acceleration term will have a negative value, indicating a reduction in speed over time.
Constant Deceleration Calculation
To calculate constant deceleration, which is a uniform decrease in velocity, we use the same kinematic equations that were used to calculate acceleration. Constant deceleration implies that the rate of change of velocity is consistent over time. We can determine the deceleration from the equation mentioned earlier by rearranging it to \( a = \frac{2(d - v_it)}{t^2} \) when time, distance, and initial velocity are known.

This formula is particularly useful for real-life scenarios, such as a car coming to a stop: in our example, given the initial speed, time until stopping, and distance covered, we can compute the constant deceleration experienced by a vehicle before impact. Positive values of \( a \) indicate acceleration, while negative values indicate deceleration, which is expected when a vehicle is coming to a halt.
Final Velocity Computation
Final velocity computation is a critical aspect when calculating an object's motion, especially in deceleration problems. It tells us the speed at which an object is traveling at a particular instant, which in this case, is right before impact. By using the first motion equation, \( v_f = v_i + at \) where \( v_f \) is the final velocity, we can easily calculate this value once we have the constant deceleration \( a \) and initial velocity \( v_i \).

After finding the constant deceleration in the previous calculation step, this formula enables us to compute the speed of the vehicle when it hits the barrier. This step is crucial for understanding the severity of an impact during a collision, and it ties in directly with practical applications in safety testing and accident reconstruction.

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Most popular questions from this chapter

A ship sets out to sail to a point \(124 \mathrm{~km}\) due north. An unexpected storm blows the ship to a point \(72.6 \mathrm{~km}\) to the north and \(31.4 \mathrm{~km}\) to the east of its starting point. How far, and in what direction, must it now sail to reach its original destination?

A dog sees a flowerpot sail up and then back down past a window \(1.1 \mathrm{~m}\) high. If the total time the pot is in sight is \(0.54 \mathrm{~s}\), find the height above the top of the window to which the pot rises.

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A ball thrown straight up takes \(2.25 \mathrm{~s}\) to reach a height of \(36.8 \mathrm{~m} .\) (a) What was its initial speed? ( \(b\) ) What is its speed at this height? ( \(c\) ) How much higher will the ball go?

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