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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
4장, 문제 4.7.107

Motion with constant acceleration The standard equation for the position s of a body moving with a constant acceleration a along a coordinate line is s = (a/2)t² + v₀t + s₀, where v₀ and s₀ are the body’s velocity and position at time t = 0. Derive this equation by solving the initial value problem
Differential equation: d²s/dt² = a
Initial conditions: ds/dt = v₀ and s = s₀ when t=0.

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Start with the given second-order differential equation: \(\frac{d^{2}s}{dt^{2}} = a\), where \(a\) is a constant acceleration.
Integrate the acceleration once with respect to time \(t\) to find the velocity \(v(t) = \frac{ds}{dt}\). This gives \(v(t) = \int a \, dt = a t + C_1\), where \(C_1\) is an integration constant.
Use the initial condition for velocity: at \(t=0\), \(v(0) = v_0\). Substitute to find \(C_1\): \(v_0 = a \cdot 0 + C_1\), so \(C_1 = v_0\). Thus, \(v(t) = a t + v_0\).
Integrate the velocity function \(v(t)\) with respect to time \(t\) to find the position function \(s(t)\): \(s(t) = \int (a t + v_0) \, dt = \frac{a}{2} t^{2} + v_0 t + C_2\), where \(C_2\) is another integration constant.
Use the initial condition for position: at \(t=0\), \(s(0) = s_0\). Substitute to find \(C_2\): \(s_0 = \frac{a}{2} \cdot 0^{2} + v_0 \cdot 0 + C_2\), so \(C_2 = s_0\). Therefore, the position function is \(s(t) = \frac{a}{2} t^{2} + v_0 t + s_0\).

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주요 개념

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

Second-Order Differential Equations

A second-order differential equation involves the second derivative of a function, representing acceleration in this context. Solving such equations requires integrating twice to find the original function, here the position s(t). Understanding how to handle these equations is essential to derive motion formulas from acceleration.
추천 영상:
07:39
Classifying Differential Equations

Initial Value Problems

An initial value problem specifies the values of a function and its derivatives at a particular point, allowing unique solutions to differential equations. Here, the initial velocity v₀ and position s₀ at time t=0 provide conditions to determine integration constants after solving the differential equation.
추천 영상:
가이드 코스
05:03
Initial Value Problems

Kinematic Equations for Constant Acceleration

Kinematic equations describe motion under constant acceleration, linking position, velocity, acceleration, and time. The given formula s = (a/2)t² + v₀t + s₀ is derived by integrating acceleration twice and applying initial conditions, illustrating the connection between calculus and classical mechanics.
추천 영상:
가이드 코스
08:14
Using The Acceleration Function
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a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values of x and y give the largest volume?

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Theory and Examples


In Exercises 53 and 54, show that the function has neither an absolute minimum nor an absolute maximum on its natural domain.


y = x¹¹ + x³ + x − 5

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10. Catching rainwater A 1125 ft^3 open-top rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.

a. If the total cost is c=5(x^2+4xy) + 10xy, what values of x and y will minimize it?

b. Give a possible scenario for the cost function in part (a).

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Absolute Extrema on Finite Closed Intervals


In Exercises 37–40, find the function’s absolute maximum and minimum values and say where they occur.


g(θ) = θ³ᐟ⁵, −32 ≤ θ ≤ 1

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Applications


A marathoner ran the 26.2-mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero.

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