Math Problem Statement

Solve it

A fighter jet lands on an aircraft carrier at a speed of 70.4 m/s and stops after 197.4 m. What is the jet's speed before covering the final 44.2 m?

Solution

The problem asks to calculate the speed of the fighter jet just before it decelerates to a complete stop after covering a distance of 44.2 m, knowing that it initially stopped after traveling 197.4 m at a speed of 70.4 m/s.

We can approach this using the equation of motion, specifically:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • vv is the final velocity (which is zero, since it stops completely after 197.4 m),
  • uu is the initial velocity (70.4 m/s),
  • aa is the acceleration (which we will find),
  • ss is the distance (197.4 m in the first case and 44.2 m in the second).

Step 1: Calculate the acceleration

For the first part of the motion (stopping over 197.4 m), the final velocity v=0v = 0 m/s, and the initial velocity u=70.4u = 70.4 m/s. Using the equation:

0=(70.4)2+2a197.40 = (70.4)^2 + 2 \cdot a \cdot 197.4

a=(70.4)22197.4a = \frac{-(70.4)^2}{2 \cdot 197.4}

Let’s calculate the value of aa:

a=4956.16394.8=12.55m/s2a = \frac{-4956.16}{394.8} = -12.55 \, \text{m/s}^2

So, the acceleration is a=12.55m/s2a = -12.55 \, \text{m/s}^2.

Step 2: Calculate the speed after traveling 44.2 m

Now, using the same equation of motion to find the speed before it covers the final 44.2 m. We need to find the initial velocity for this part. The equation becomes:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • s=44.2ms = 44.2 \, \text{m},
  • a=12.55m/s2a = -12.55 \, \text{m/s}^2,
  • uu is what we need to find (the initial speed before the deceleration),
  • v=0m/sv = 0 \, \text{m/s} (since it stops).

Rearranging the equation:

0=u2+2(12.55)(44.2)0 = u^2 + 2(-12.55)(44.2)

u2=2(12.55)44.2=1108.82u^2 = -2 \cdot (-12.55) \cdot 44.2 = 1108.82

u=1108.82=33.3m/su = \sqrt{1108.82} = 33.3 \, \text{m/s}

Thus, the speed of the jet before traveling the last 44.2 m is approximately 33.3 m/s.


Would you like further clarification or details on this solution? Here are some related questions for further practice:

  1. What is the total time taken for the jet to come to a stop from 70.4 m/s?
  2. What would the deceleration be if the jet stopped over a distance of 300 meters instead?
  3. How would the initial speed change if the stopping distance were doubled?
  4. How much time does the jet take to reduce its speed from 33.3 m/s to zero over the final 44.2 meters?
  5. Can you derive a formula for the distance covered during a deceleration from any speed to rest?

Tip: Always keep track of your signs when dealing with acceleration and deceleration, as confusion can lead to errors!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Equations of Motion

Formulas

v^2 = u^2 + 2as
a = (v^2 - u^2) / (2s)

Theorems

Kinematic equations for uniformly accelerated motion

Suitable Grade Level

High School (Grades 10-12)