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An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1 , the magnitude of the car's acceleration is \(3.0 \mathrm{~m} / \mathrm{s}^{2}\). The magnitude of the car's velocity at the end of stage 2 is \(2.5\) times greater than it is at the end of stage 1 . Find the magnitude of the acceleration in stage 2 .

Short Answer

Expert verified
The acceleration in stage 2 is \\(4.5 \, \text{m/s}^2\\).

Step by step solution

01

Identify Known Variables

We have an automobile starting from rest, which means the initial velocity \(v_0 = 0\). The acceleration during stage 1 is \(a_1 = 3.0 \, \text{m/s}^2\). At the end of stage 2, the final velocity is 2.5 times the velocity at the end of stage 1. We are tasked to find the acceleration during stage 2.
02

Calculate Velocity at End of Stage 1

Assume that both stages take the same amount of time \(t\). Using the equation of motion \(v = u + at\), where \(u\) is the initial velocity, the velocity at the end of stage 1 \(v_1\) can be calculated as:\[v_1 = 0 + a_1 \cdot t = 3.0 \cdot t \, \text{m/s}\]
03

Calculate Velocity at End of Stage 2

At the end of stage 2, the velocity \(v_2\) is 2.5 times \(v_1\):\[v_2 = 2.5 \cdot v_1 = 2.5 \cdot (3.0 \cdot t) = 7.5 \cdot t \, \text{m/s}\]
04

Determine Acceleration for Stage 2

Using the equation \(v = u + at\) again for stage 2, where the initial velocity is \(v_1\) and the final velocity is \(v_2\), we can find the acceleration \(a_2\):\[v_2 = v_1 + a_2 \cdot t\]Solving for \(a_2\):\[7.5t = 3.0t + a_2 \cdot t \4.5t = a_2 \cdot t \a_2 = 4.5 \, \text{m/s}^2\]

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Key Concepts

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

acceleration
In kinematics, acceleration is a crucial concept that explains the rate at which an object's velocity changes over time. For any moving vehicle, it's important to understand how acceleration contributes to its motion. In this particular exercise, we have a car that accelerates in two stages. The acceleration in the first stage is noted to be 3.0 m/s². But what does this mean?
  • Acceleration is the change in velocity per unit of time. Here, it means that for every second, the car's velocity increases by 3 m/s.
  • It is a vector quantity, which means it has both magnitude and direction.
In practical terms, you can think of pushing the gas pedal in a car. The harder and longer you push, the quicker the car speeds up. In the exercise, determining the second stage's acceleration required understanding how much faster the car needed to get compared to the first stage. Thus, stage 2's acceleration was calculated as 4.5 m/s², showing how the car's speed was increasing more rapidly in the final leg of its journey.
velocity
Velocity in physics refers to the speed of an object in a given direction. It's more than just how fast something is moving; it's also about where it's headed. In this exercise:
  • Velocity isn't just speed; it's speed with direction. So a car receding at 30 m/s has a velocity of 30 m/s in the opposite direction.
  • Initial velocity is what the car has before it starts its journey, which, in this case, was 0 since it started from rest.
To determine the stages of the car's velocity, we applied equations that rely on the relationship between final and initial velocities, time, and acceleration. By the end of stage 1, the car achieved a velocity of 3t m/s, meaning how fast it was moving depended directly on how long the stage lasted. By the end of stage 2, our task was to find 2.5 times this velocity, leading to a final speed calculation of 7.5t m/s. Thus, the calculation shows the car significantly picking up speed between these stages.
equations of motion
In the realm of kinematics, equations of motion allow us to foresee how an object will behave when moving under a uniform acceleration. These equations bridge initial conditions with final outcomes by considering time and acceleration.
  • The primary equation used frequently here is: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
  • This formula helps predict velocity at any given moment based on its start and the acceleration it experiences.
In the exercise, we broke these down into stages using given accelerations and time to calculate final velocities. By identifying known variables, applying the formula twice for different stages, and solving for new accelerations, these equations methodically guide us through solving complex motion scenarios, as demonstrated by determining the car's acceleration in stage 2 using known and calculated values from stage 1.

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Most popular questions from this chapter

A sprinter accelerates from rest to a top speed with an acceleration whose magnitude is \(3.80 \mathrm{~m} / \mathrm{s}^{2}\). After achieving top speed, he runs the remainder of the race without speeding up or slowing down. The total race is fifty meters long. If the total race is run in \(7.88 \mathrm{~s}\), how far does he run during the acceleration phase?

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed of \(1.8 \mathrm{~m} / \mathrm{s}\) from a \(3.0-\mathrm{m}\) board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is \(y=-3.0 \mathrm{~m}\) (measured from the board), assuming that the downward direction is chosen as the negative direction. \(]\) (b) What is the highest point he reaches above the water?

In preparation for this problem, review Conceptual Example 7 . From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The initial speed of the pebble is \(9.0 \mathrm{~m} / \mathrm{s}\). (a) What is the acceleration (magnitude and direction) of the pebble during the downward motion? Is the pebble decelerating? Explain. (b) After \(0.50 \mathrm{~s}\), how far beneath the cliff top is the pebble?

A woman and her dog are out for a morning run to the river, which is located \(4.0 \mathrm{~km}\) away. The woman runs at \(2.5 \mathrm{~m} / \mathrm{s}\) in a straight line. The dog is unleashed and runs back and forth at \(4.5 \mathrm{~m} / \mathrm{s}\) between his owner and the river, until the woman reaches the river. What is the total distance run by the dog?

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{~m} / \mathrm{s},\) and during a 5.0 -s interval, it changes to a final speed of (a) \(29.0 \mathrm{~m} / \mathrm{s}\) and ( b) \(23.0 \mathrm{~m} / \mathrm{s}\). In each case, find the acceleration (magnitude and algebraic sign) and state whether or not the car is decelerating.

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