Math Problem Statement

You are in a car that’d traveling at a velocity of 12.24 meters a second. You receive a call that you need to be at an important event in 25.23 minutes. You start to accelerate at a rate of 0.525 meters a second squared. How far would you travel in 25.23 minutes?

Solution

To calculate how far you would travel given this scenario, we need to break the problem into steps and use the basic kinematic equations of motion. Here's a step-by-step breakdown:

Given:

  • Initial velocity (v0v_0) = 12.24 m/s
  • Time (tt) = 25.23 minutes = 25.23×60=1513.825.23 \times 60 = 1513.8 seconds
  • Acceleration (aa) = 0.525 m/s²

Formula to Use:

The equation of motion for distance traveled under constant acceleration is:

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

Where:

  • dd is the total distance traveled.
  • v0v_0 is the initial velocity.
  • tt is the time.
  • aa is the acceleration.

Calculation:

Substitute the values into the formula.

d=12.24×1513.8+12×0.525×(1513.8)2d = 12.24 \times 1513.8 + \frac{1}{2} \times 0.525 \times (1513.8)^2

Let's compute this.The total distance you would travel in 25.23 minutes, while accelerating at 0.525 m/s², is approximately 620,071.4 meters (or about 620 kilometers).

Would you like further details on any of the steps, or do you have other questions?

Here are some related questions:

  1. How would the distance change if the initial velocity were higher?
  2. What if the acceleration was negative (deceleration)?
  3. How long would it take to reach a certain speed given this acceleration?
  4. How much distance is traveled in the first 5 minutes?
  5. How would the result change if the acceleration was applied after a delay?

Tip: Always ensure the units of time are consistent (seconds, minutes, etc.) when solving kinematics problems.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Constant Acceleration
Time-Distance Relationship

Formulas

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

Theorems

Kinematic Equations for Constant Acceleration

Suitable Grade Level

Grades 10-12