Questions: What is the displacement Δx of the particle? Express your answer in meters. Δx=□ √[□]□ A Σ φ+m

Solution Steps

The graph shows the velocity \(v_x\) of a particle as a function of time \(t\). The displacement \(\Delta x\) is the area under the velocity-time graph.

The area under the graph can be divided into a triangle, a rectangle, and another triangle.

The first triangle spans from \(t=0 \, \text{s}\) to \(t=20 \, \text{s}\) and has a height of \(2.0 \, \text{m/s} - 0.5 \, \text{m/s} = 1.5 \, \text{m/s}\). Its area is:

\(A_1 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot (20 \, \text{s}) \cdot (1.5 \, \text{m/s}) = 15 \, \text{m}\)

The rectangle spans from \(t=20 \, \text{s}\) to \(t=45 \, \text{s}\) and has a height of \(2.0 \, \text{m/s}\). Its area is:

\(A_2 = \text{length} \cdot \text{width} = (45 \, \text{s} - 20 \, \text{s}) \cdot (2.0 \, \text{m/s}) = 25 \, \text{s} \cdot 2.0 \, \text{m/s} = 50 \, \text{m}\)

The second triangle spans from \(t=45 \, \text{s}\) to \(t=50 \, \text{s}\) and has a height of \(2.0 \, \text{m/s}\). Its area is:

\(A_3 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot (50 \, \text{s} - 45 \, \text{s}) \cdot (2.0 \, \text{m/s}) = \frac{1}{2} \cdot 5 \, \text{s} \cdot 2.0 \, \text{m/s} = 5 \, \text{m}\)

The total displacement is the sum of the areas:

\(\Delta x = A_1 + A_2 + A_3 = 15 \, \text{m} + 50 \, \text{m} + 5 \, \text{m} = 70 \, \text{m}\)

Final Answer