A particle moving along the x-axis has its veocity described by the function vx = 2t2 m/s, where t is in s. itsinitial position is x0 = 1 m at t0 = 0 s. At t = 1 s, what are the particle's position?

Ch 02: Kinematics in One Dimension

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 02: Kinematics in One Dimension

문제 34a

2장, 문제 34a

A particle moving along the x-axis has its veocity described by the function v_x = 2t² m/s, where t is in s. itsinitial position is x₀ = 1 m at t₀ = 0 s. At t = 1 s, what are the particle's position?

검증된 단계별 안내

Step 1: Understand the problem. The velocity of the particle is given as a function of time, \( v_x = 2t^2 \) (in m/s). The initial position of the particle is \( x_0 = 1 \) m at \( t_0 = 0 \) s. We need to find the position of the particle at \( t = 1 \) s.

Step 2: Recall the relationship between velocity and position. The position \( x(t) \) can be found by integrating the velocity function \( v_x(t) \) with respect to time: \( x(t) = \int v_x(t) \, dt + C \), where \( C \) is the constant of integration.

Step 3: Perform the integration. Substitute \( v_x(t) = 2t^2 \) into the integral: \( x(t) = \int 2t^2 \, dt + C \). The integral of \( 2t^2 \) is \( \frac{2t^3}{3} \), so \( x(t) = \frac{2t^3}{3} + C \).

Step 4: Determine the constant of integration \( C \). Use the initial condition \( x(0) = 1 \): \( 1 = \frac{2(0)^3}{3} + C \). This simplifies to \( C = 1 \). Therefore, the position function becomes \( x(t) = \frac{2t^3}{3} + 1 \).

Step 5: Calculate the position at \( t = 1 \) s. Substitute \( t = 1 \) into the position function: \( x(1) = \frac{2(1)^3}{3} + 1 \). Simplify this expression to find the position of the particle at \( t = 1 \) s.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Velocity

Velocity is the rate of change of an object's position with respect to time. It is a vector quantity, meaning it has both magnitude and direction. In this case, the velocity function vx = 2t^2 indicates that the velocity of the particle changes with time, specifically increasing as the square of time. Understanding how to interpret and manipulate this function is crucial for determining the particle's position.

Integration

Integration is a fundamental concept in calculus used to find the accumulated value of a quantity over an interval. In the context of motion, integrating the velocity function with respect to time gives the position function. This process allows us to determine how far the particle has traveled from its initial position over a given time period, which is essential for solving the problem.

Initial Conditions

Initial conditions refer to the values of a system's variables at the starting point of observation. In this problem, the initial position x0 = 1 m at t0 = 0 s provides a reference point for calculating the particle's position at later times. Incorporating initial conditions is vital for accurately determining the final position after applying the results of integration to the velocity function.