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The particle *p* travels along a path as illustrated in Fig. 1. The initial speed of the particle is *v*=1*m*/*s* at the position O, and the speed increases by a=0.1*m*/*s*2. Determine the position of the particle on the x and y axes at the instant 5 seconds.

Given the problem, we need to determine the position of the particle on the \( x \) and \( y \) axes at the instant \( t = 5 \) seconds.

### Problem Breakdown:

- **Initial Speed:** \( v_0 = 1 \, \text{m/s} \)
- **Constant Acceleration:** \( a = 0.1 \, \text{m/s}^2 \)
- **Time:** \( t = 5 \) seconds
- **Path Equation:** \( y = \frac{1}{2}x^2 \)

### Step 1: Determine the velocity at \( t = 5 \) seconds

Using the kinematic equation for velocity:
\[
v(t) = v_0 + a \cdot t
\]
Substituting the given values:
\[
v(5) = 1 + 0.1 \times 5 = 1 + 0.5 = 1.5 \, \text{m/s}
\]

### Step 2: Determine the distance traveled along the curve (x-axis) after 5 seconds

The distance traveled under constant acceleration is given by:
\[
s = v_0 t + \frac{1}{2} a t^2
\]
Substituting the values:
\[
s = 1 \times 5 + \frac{1}{2} \times 0.1 \times 5^2 = 5 + \frac{1}{2} \times 0.1 \times 25 = 5 + 1.25 = 6.25 \, \text{meters}
\]
However, \( s \) represents the distance along the path, and since \( x \) and \( y \) are related through the curve, we can find \( x \) by integrating.

### Step 3: Determine the position of the particle

Assuming the motion is along the path given by \( y = \frac{1}{2}x^2 \), we need to find the \( x \) and \( y \) positions.

To find \( x(t) \) and \( y(t) \), the integral of velocity must consider the path's constraints:
\[
x(t) = \int_0^t v(t) \, dt
\]

Given that the speed \( v \) is non-linear (due to acceleration), we'd expect that the distance along the curve is also nonlinear, necessitating the solving of the integral considering the curve's shape and re-integrating the equation.

If the motion is uniform along the curve, we can use:
\[
x = \sqrt{\frac{2s}{1}} \text{ (where \( y = \frac{1}{2}x^2 \))}
\]
Since \( s = 6.25 \) meters and \( y = \frac{1}{2}x^2 \):

We approximate the values:
\[
x = \sqrt{2 \times 6.25} \approx \sqrt{12.5} \approx 3.54 \text{ meters}
\]

Then, \( y = \frac{1}{2} \times (3.54)^2 \approx 6.25 \times 0.5 = 3.12 \text{ meters} \).

### Conclusion

At \( t = 5 \) seconds, the particle will approximately be at:
- \( x \approx 3.54 \) meters
- \( y \approx 3.12 \) meters

Would you like more details on any part of this solution, or do you have any questions? Here are some follow-up questions to consider:

1. How would the solution change if the acceleration was variable instead of constant?
2. What if the initial velocity was zero?
3. How does the shape of the path influence the particle's speed at any point?
4. What if the path equation were different (e.g., \( y = x^3 \))? How would you approach the problem?
5. How can the concept of work-energy theorem be applied to verify the results?
6. What would happen to the position if the acceleration were in the opposite direction?
7. Can we determine the exact velocity vector at \( t = 5 \) seconds?
8. How would friction or resistance affect the particle’s motion?

**Tip:** Always check the relationship between position, velocity, and acceleration for motion along a curved path, as they can be non-linear due to the geometry of the path.

Learn how to determine the position of a particle on the x and y axes after 5 seconds with a detailed mathematical solution. This problem involves kinematics, calculus, and the integration of non-linear velocity over a path defined by y = 0.5x^2. Suitable for advanced high school students.

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