Math Problem Statement

Problem 3: (13% of Assignment Value)  An airplane covers a straight-line distance of 8.27 km in 31.1 s, during which time it has a constant forward acceleration of 3.82 m/s2.

huy.tran8@mavs.uta.edu  5T88-3F-BF-4A-B93C-53404

      Part (a)   

What is the speed of the airplane, in m/s, at the beginning of the 31.1 s? 

v0 = 

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tan()

cotan()

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acos()

atan()

acotan()

sinh()

cosh()

tanh()

cotanh()

Degrees  Radians

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huy.tran8@mavs.uta.edu

5T88-3F-BF-4A-B93C-53404

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      Part (b)   

What is the speed of the airplane, in m/s, at the end of the 31.1

Solution

To solve this problem, we will use the basic kinematic equations for constant acceleration. Let's break it down step by step for both parts (a) and (b).

Given Data:

  • Distance covered: d=8.27km=8270md = 8.27 \, \text{km} = 8270 \, \text{m}
  • Time: t=31.1st = 31.1 \, \text{s}
  • Constant acceleration: a=3.82m/s2a = 3.82 \, \text{m/s}^2

Part (a): Initial Speed v0v_0

We need to find the initial speed v0v_0 of the airplane at the start of the time period. For this, we can use the following kinematic equation:

d=v0t+12at2d = v_0 t + \frac{1}{2} a t^2

Where:

  • dd is the distance,
  • v0v_0 is the initial speed (what we need to find),
  • tt is the time,
  • aa is the acceleration.

Rearranging the equation to solve for v0v_0:

v0=d12at2tv_0 = \frac{d - \frac{1}{2} a t^2}{t}

Now, substituting the known values:

v0=827012×3.82×(31.1)231.1v_0 = \frac{8270 - \frac{1}{2} \times 3.82 \times (31.1)^2}{31.1}

Let me calculate that for you.

Part (b): Final Speed vfv_f

To find the final speed vfv_f of the airplane at the end of the 31.1 seconds, we can use another kinematic equation:

vf=v0+atv_f = v_0 + a t

After calculating v0v_0 from part (a), we will substitute it into this equation along with the known acceleration and time to find the final speed.

Let me calculate both v0v_0 and vfv_f.### Results:

  • Initial speed of the airplane at the beginning of 31.1 seconds: v0=206.52m/sv_0 = 206.52 \, \text{m/s} (rounded to two decimal places).
  • Final speed of the airplane at the end of 31.1 seconds: vf=325.32m/sv_f = 325.32 \, \text{m/s} (rounded to two decimal places).

Would you like more details on the calculations or explanations?

Here are some related questions to explore:

  1. What is the total distance covered by the airplane if the acceleration continues for another 10 seconds?
  2. How long would it take the airplane to reach a speed of 500 m/s under the same acceleration?
  3. What would be the airplane's speed after 20 seconds if it started from rest?
  4. How would the result change if the acceleration were reduced to 2 m/s²?
  5. What would happen to the distance covered if the acceleration were doubled?

Tip: For constant acceleration, using the right kinematic equation helps solve for unknowns like speed, time, or distance based on what is provided.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Constant Acceleration
Distance-Time Relationships

Formulas

d = v_0 * t + (1/2) * a * t²
v_f = v_0 + a * t

Theorems

Kinematic Equations for Constant Acceleration

Suitable Grade Level

Grades 11-12 (Physics)