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Two rockets having the same acceleration start from rest, but rocket \(A\) travels for twice as much time as rocket \(B\) . If rocket \(A\) goes a distance of \(250 \mathrm{km},\) how far will rocket \(B\) go? (b) If rocket \(A\) reaches a speed of \(350 \mathrm{m} / \mathrm{s},\) what speed will rocket \(B\) reach?

Short Answer

Expert verified
Rocket B travels 62.5 km and reaches a speed of 175 m/s.

Step by step solution

01

Understand the variables

Identify the variables from the problem.Let the acceleration common to both rockets be \(a\), let the time taken for rocket \(A\) be \(2t\) and the time for rocket \(B\) be \(t\). The distance rocket \(A\) travels is \(250\, \text{km}\), which is \(250,000\, \text{m}\).
02

Use the equation for distance

Recall the equation for distance traveled under constant acceleration from rest: \(d = \frac{1}{2} a t^2\).For rocket \(A\), substitute with \(2t\):\[250,000 = \frac{1}{2} a (2t)^2 = 2at^2\]Solve for \(at^2\), we get \(at^2 = 125,000\).
03

Find distance for rocket B

Use the same distance formula for rocket \(B\): \(d = \frac{1}{2} a t^2\).Since \(at^2 = 125,000\), we have:\[d_B = \frac{1}{2} \times 125,000 = 62,500\, \text{m}\] Convert to kilometers: \(62.5\, \text{km}\).
04

Use the equation for final velocity

Recall the equation for final velocity under constant acceleration: \(v = at\).For rocket \(A\), since \(v_A = 350\, \text{m/s}\), and \(at = 350 / 2\), solve for \(at\):\[at = 175\]
05

Calculate velocity for rocket B

Use \(v_B = at\) for rocket \(B\).Thus, from \(at = 175\), we get:\[v_B = 175\, \text{m/s}\]

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object speeds up. It is defined as the rate of change of velocity per unit time. When a rocket speeds up from rest, it experiences a constant acceleration if this rate doesn't change over time. To calculate acceleration, we use the formula:\[a = \frac{\Delta v}{\Delta t}\]where \(a\) is acceleration, \(\Delta v\) is the change in velocity, and \(\Delta t\) is the change in time. Acceleration can increase or decrease speed and can be measured in meters per second squared (\(m/s^2\)).When dealing with two rockets having the same acceleration, as in this problem, it means both rockets speed up at the same rate. This is crucial in solving physics problems as it provides a consistent framework for calculations.
Distance Calculation
Distance calculation during constant acceleration can be a bit intricate but is manageable with the right formula. When an object starts from rest like our rockets, the equation to find the distance \(d\) covered is:\[d = \frac{1}{2} a t^2\]This formula works under the assumption that acceleration, \(a\), is constant and the initial velocity is zero. In the exercise, rocket A travels 250 km, and using this distance formula, we determine the shared variable \(at^2\) as 125,000.
  • For Rocket A, we substitute the time as \(2t\) to calculate its specific travel distance.
  • For Rocket B, simply use the same formula with time \(t\) to find its distance.
This helps calculate that Rocket B covers a distance of 62.5 km. All calculations must be made carefully to ensure problems about motion are solved correctly.
Final Velocity
The final velocity of an object under constant acceleration can be determined using the equation:\[v = at\]This represents velocity (\(v\)) in terms of acceleration (\(a\)) and time (\(t\)). It's an important formula as it helps understand how fast an object will be moving after a certain period of constant acceleration.
  • For Rocket A, the final velocity is given as 350 m/s. By substituting \(at = 175\), we can determine this value for Rocket B as well.
  • Rocket B's resulting velocity turns out to be 175 m/s by using the same process.
This concept is a key part of kinematics in physics, allowing predictions of motion and speed under specified conditions.
Constant Acceleration
Constant acceleration is a fundamental concept in physics that simplifies the calculation of motion. It occurs when an object's velocity changes at a consistent rate over time. This is especially significant when analyzing motion in straight lines or freefall conditions.
  • In scenarios like this exercise, both rockets experience the same constant acceleration. This uniform rate means calculations for time, distance, and velocity follow predictable formulas.
  • The predictability of constant acceleration allows use of the classic equations of motion, making physics problems more approachable.
Constant acceleration is not just a theoretical part of physics; it applies to real-world situations like cars moving at a steady speed increase or even objects in gravitational fields. By understanding this, solving problems becomes a logical process based on firm principles.

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Most popular questions from this chapter

Airplane \(A,\) starting from rest with constant acceleration, requires a runway 500 \(\mathrm{m}\) long to become airborne. Airplane \(B\) requires a takeoff speed twice as great as that of airplane \(A\) but has the same acceleration, and both planes start from rest. (a) How long must the runway be for airplane \(B ?\) (b) If airplane \(A\) takes time \(T\) to travel the length of its runway, how long (in terms of \(T )\) will airplane \(B\) take to travel the length of its runway?

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