Math Problem Statement

The particle p travels along a path as illustrated in Fig. 1. The initial speed of the particle is 𝑣(𝑑)=1π‘š 𝑠. at the position O, and the speed increases by 𝑣˙(𝑑)=0.1π‘š/𝑠2. Determine the acceleration of the particle at the instant 5 seconds.

Solution

To determine the acceleration of the particle at the instant 5 seconds, we can proceed as follows:

Step 1: Identify Given Information

  • Initial speed at position OO: v(0)=1 m/sv(0) = 1 \, \text{m/s}
  • Rate of change of speed (which is acceleration): vΛ™(t)=0.1 m/s2\dot{v}(t) = 0.1 \, \text{m/s}^2

Step 2: Find the Speed at t=5t = 5 Seconds

Since the speed increases at a constant rate, we can express the speed at any time tt using: v(t)=v(0)+vΛ™(t)β‹…tv(t) = v(0) + \dot{v}(t) \cdot t Substituting the given values: v(5)=1 m/s+(0.1 m/s2)β‹…5 s=1+0.5=1.5 m/sv(5) = 1 \, \text{m/s} + (0.1 \, \text{m/s}^2) \cdot 5 \, \text{s} = 1 + 0.5 = 1.5 \, \text{m/s}

Step 3: Determine the Acceleration at t=5t = 5 Seconds

The total acceleration a(t)a(t) of the particle can be broken down into two components:

  1. Tangential acceleration (ata_t): This is the rate of change of the particle's speed along its path and is equal to vΛ™(t)\dot{v}(t).
  2. Normal (centripetal) acceleration (ana_n): This is due to the change in direction of the velocity and is given by: an=v(t)2ρa_n = \frac{v(t)^2}{\rho} where ρ\rho is the radius of curvature of the path at the point of interest.

Given:

  • Tangential acceleration: at=vΛ™(t)=0.1 m/s2a_t = \dot{v}(t) = 0.1 \, \text{m/s}^2
  • The velocity at t=5t = 5 seconds: v(5)=1.5 m/sv(5) = 1.5 \, \text{m/s}

Step 4: Calculate the Total Acceleration

The total acceleration is the vector sum of the tangential and normal accelerations: a(t)=at2+an2a(t) = \sqrt{a_t^2 + a_n^2} However, the radius of curvature ρ\rho is not provided directly. If the radius of curvature or further details about the path are not given, the tangential acceleration would be considered the primary component of acceleration at that instant.

Thus, the total acceleration at t=5t = 5 seconds is primarily: a(5)=0.1 m/s2a(5) = 0.1 \, \text{m/s}^2 assuming the normal component is negligible or not calculable with the provided information.

Would you like further details on calculating the radius of curvature or any other aspects of this problem?


Further Questions to Consider:

  1. How do you calculate the radius of curvature for a specific point on a curve?
  2. What is the relationship between speed, velocity, and acceleration in curvilinear motion?
  3. How does the tangential acceleration change with varying speed?
  4. What additional information would allow you to calculate the normal acceleration?
  5. How do normal and tangential accelerations affect the motion of the particle?
  6. How is the path of the particle influenced by its initial velocity?
  7. What is the significance of the initial conditions given at position OO?
  8. How would the acceleration change if the rate of speed increase vΛ™(t)\dot{v}(t) were not constant?

Tip: Always check if the normal (centripetal) acceleration component is required for a complete solution in curvilinear motion problems.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Acceleration
Curvilinear Motion

Formulas

Speed calculation formula
Acceleration components formula

Theorems

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Suitable Grade Level

Advanced High School