Question

Displacement from velocity A particle moves along a line with a velocity given by v(t) = 5 sin πt, starting with an initial position s(0) = 0 . Find the displacement of the particle between t = 0 and t = 2 , which is given by s(t) = ∫₀² v(t) dt . Find the distance traveled by the particle during this interval, which is ∫₀² |v(t)| dt .

Accepted Answer

Step 1: Understand the problem. The displacement of the particle is calculated using the definite integral of the velocity function v(t) over the interval [0, 2]. The distance traveled is calculated using the definite integral of the absolute value of the velocity function |v(t)| over the same interval. Step 2: Write the displacement formula. The displacement is given by s(t) = ∫₀² v(t) dt. Substitute v(t) = 5 sin(πt) into the integral: s(t) = ∫₀² 5 sin(πt) dt. Step 3: Solve the integral for displacement. Use the integral rule for sine: ∫ sin(ax) dx = -(1/a) cos(ax) + C. Here, a = π, so the integral becomes ∫₀² 5 sin(πt) dt = -5/π [cos(πt)] evaluated from t = 0 to t = 2. Step 4: Write the formula for distance traveled. The distance is given by ∫₀² |v(t)| dt. Since v(t) = 5 sin(πt), the absolute value |v(t)| must be considered. Analyze the behavior of sin(πt) over [0, 2] to determine where it is positive or negative, and split the integral accordingly. Step 5: Solve the integral for distance. Break the integral into intervals where sin(πt) is positive and negative. For t ∈ [0, 1], sin(πt) ≥ 0, so |v(t)| = 5 sin(πt). For t ∈ [1, 2], sin(πt) \u003c 0, so |v(t)| = -5 sin(πt). Compute ∫₀¹ 5 sin(πt) dt and ∫₁² -5 sin(πt) dt separately, then add the results to find the total distance traveled.

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Key Concepts

Velocity and Displacement

Definite Integral

Absolute Value of Velocity