Question

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.

Accepted Answer

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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Key Concepts

Second-Order Differential Equations

Initial Value Problems

Kinematic Equations for Constant Acceleration