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At noon a car starts from rest at point \(A\) and proceeds with constant acceleration along a straight road toward point \(C\), 35 miles away. If the constantly accelerated car arrives at \(C\) with a velocity of \(60 \mathrm{mi} / \mathrm{h}\), at what time does it arrive at \(C ?\)

Short Answer

Expert verified
The car arrives at point C around 1:10 PM.

Step by step solution

01

Understand the Given Information

The car starts from rest, meaning its initial velocity is 0 mi/h. It accelerates constantly towards point C, which is 35 miles away. By the time it reaches point C, its velocity is 60 mi/h.
02

Identify the Relevant Equation

We need to find the time it takes for the car to reach point C. The equation of motion that relates initial velocity \(u\), final velocity \(v\), acceleration \(a\), and distance \(s\) is:\[ v^2 = u^2 + 2as \].Substitute the given values \(u = 0\), \(v = 60\) mi/h, and \(s = 35\) miles.
03

Restate the Equation in Terms of Acceleration

The equation \(60^2 = 0^2 + 2a(35)\) simplifies to:\[ 3600 = 70a \].Solve for \(a\) by dividing both sides by 70: \[ a = \frac{3600}{70} \].
04

Calculate the Acceleration

Solve the equation to find the acceleration \(a\):\[ a = 51.43 \text{ mi/h}^2 \].Now we know the car's constant acceleration.
05

Use the Time Equation

With acceleration determined, use the equation \(v = u + at\) to find time \(t\). Rearrange it to solve for \(t\):\[ 60 = 0 + (51.43)t \].
06

Calculate the Time

Solve for \(t\):\[ t = \frac{60}{51.43} \approx 1.167 \text{ hours} \]. Convert this to minutes by multiplying by 60: \[ t \approx 70 \text{ minutes} \].
07

Determine the Arrival Time

Since the car left at noon and it takes approximately 70 minutes to reach point C, add this duration to 12:00 PM. The car arrives at point C around 1:10 PM.

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Key Concepts

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

Constant Acceleration
In physics, the concept of constant acceleration is crucial when analyzing motion. Constant acceleration means that the velocity of an object changes at a steady rate over time.
For example, in our car scenario from the exercise, it starts from rest and progressively speeds up at a constant rate until it reaches its designated velocity. This steady change in speed means that with each passing moment, the car's velocity increases by equal amounts. Using constant acceleration helps simplify the calculations as it allows us to use certain equations of motion that are derived for bodies moving with unchanging acceleration. In situations where the acceleration isn't constant, these equations cannot be applied directly without modification.
Understanding constant acceleration means grasping two key points:
  • The object's speed is changing continuously and uniformly.
  • This change in speed takes place over a period of time.
Handling constant acceleration carefully helps in making accurate predictions about future motion, such as when and where the car will reach its destination.
Equations of Motion
Equations of motion are fundamental tools in physics used to predict an object's future position and velocity. They are particularly useful when dealing with objects moving under constant acceleration. In this exercise, the car moves with constant acceleration to cover a distance.One critical equation is:\[ v^2 = u^2 + 2as \]This equation relates the final velocity (\( v \)) of the moving object to its initial velocity (\( u \)), the acceleration (\( a \)), and the distance covered (\( s \)).
In our example, the initial velocity (\( u \)) is 0 because the car starts from rest. There's also the time-dependent equation:\[ v = u + at \]Each of these equations provides a way to calculate missing pieces of information like acceleration or time taken when a body moves under constant acceleration. By substituting known values into these equations, as we did when calculating the car's journey, we gain valuable insight into the dynamics of the motion.
Velocity Calculation
Velocity calculation is an essential part of understanding motion. In this case, the car's final velocity as it arrives at point \( C \) is 60 mi/h.
Calculating velocity involves understanding how speed and direction combine to describe motion fully:Velocity, a vector quantity, includes both the magnitude (or speed) and the direction of an object. While speed is simply how fast something is moving, velocity tells you how fast and in what direction.For our exercise, the car's velocity changes from 0 to 60 mi/h under constant acceleration. This allows us to use equations like:
  • \[ v = u + at \]
The initial velocity \( u \) is zero because the car starts from rest. By using the equation and the final velocity \( v \), we determined the acceleration \( a \). The next step using this acceleration was to calculate the time \( t \) taken.Study and calculation of velocity help us answer questions about "how fast" an object moves and "how soon" it will reach its destination. Understanding velocity is key to solving numerous kinematic problems in physics.

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