Math Problem Statement

  1. A certain airplane has an acceleration of 15.0 m/s2. a) How fast will it be moving 2.5 s after it starts down the runway? b) How far down the runway will it travel during the 2.5 s? c) Minimum take-off speed is 60.0 m/s. How long must the runway be?

Solution

We are given that the airplane has an acceleration of a=15.0m/s2a = 15.0 \, \text{m/s}^2, and we need to find:

a) How fast will it be moving 2.5 s after it starts down the runway?

We can use the equation for velocity under constant acceleration: v=u+atv = u + at Where:

  • vv is the final velocity
  • uu is the initial velocity (since it starts from rest, u=0u = 0)
  • a=15.0m/s2a = 15.0 \, \text{m/s}^2 is the acceleration
  • t=2.5secondst = 2.5 \, \text{seconds}

Substituting the values: v=0+(15.0m/s2)(2.5s)=37.5m/sv = 0 + (15.0 \, \text{m/s}^2)(2.5 \, \text{s}) = 37.5 \, \text{m/s} So, the airplane will be moving at 37.5 m/s after 2.5 seconds.

b) How far down the runway will it travel during the 2.5 s?

For distance, we use the equation: s=ut+12at2s = ut + \frac{1}{2}at^2 Where:

  • ss is the distance traveled
  • u=0m/su = 0 \, \text{m/s}
  • a=15.0m/s2a = 15.0 \, \text{m/s}^2
  • t=2.5secondst = 2.5 \, \text{seconds}

Substituting the values: s=(0)(2.5)+12(15.0)(2.5)2=12(15.0)(6.25)=46.875ms = (0)(2.5) + \frac{1}{2}(15.0)(2.5)^2 = \frac{1}{2}(15.0)(6.25) = 46.875 \, \text{m} So, the airplane will travel 46.9 meters down the runway in 2.5 seconds.

c) How long must the runway be if the minimum take-off speed is 60.0 m/s?

We use the same velocity equation: v=u+atv = u + at Rearranging it to solve for time tt: t=vua=60.0015.0=4.0secondst = \frac{v - u}{a} = \frac{60.0 - 0}{15.0} = 4.0 \, \text{seconds} Now, to find the required runway length, use the distance formula: s=ut+12at2s = ut + \frac{1}{2}at^2 Substitute the values: s=(0)(4.0)+12(15.0)(4.0)2=12(15.0)(16)=120ms = (0)(4.0) + \frac{1}{2}(15.0)(4.0)^2 = \frac{1}{2}(15.0)(16) = 120 \, \text{m} Thus, the minimum runway length must be 120 meters.

Summary:

a) Final speed after 2.5 s: 37.5 m/s
b) Distance traveled in 2.5 s: 46.9 m
c) Minimum runway length for take-off: 120 m

Would you like more details or have any questions?

Here are 5 related questions to explore further:

  1. What happens to the take-off time if the acceleration is increased?
  2. How would the distance traveled change if the airplane accelerated for 5 seconds?
  3. What role does air resistance play in the actual take-off process?
  4. How would you calculate the energy required for take-off?
  5. Can you determine the force acting on the airplane during acceleration?

Tip: Always double-check the units in your calculations to ensure consistency!

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Math Problem Analysis

Mathematical Concepts

Kinematics
Uniform Acceleration
Distance-Time-Velocity Relationships

Formulas

v = u + at
s = ut + (1/2)at^2
t = (v - u) / a

Theorems

Equations of motion under constant acceleration

Suitable Grade Level

Grades 9-12