Question

36 5ep construct an explanation go online to find a graph for accelerated motion that is not constant. use the graph to explain why the displacement equation of motion is not appropriate for analyzing accelerated motion that is not constant.

Explanation:

1. Find a non - constant acceleration graph: Search online for a velocity - time graph where the slope (acceleration) changes, e.g., a velocity - time graph with a curved line (like a parabola opening upwards/downwards or a more complex curve). For example, consider a velocity - time graph of an object under a force that varies with time, so acceleration \(a=\frac{F}{m}\) (Newton's second law) also varies. If the force is \(F(t)=kt^{2}\) (where \(k\) is a constant and \(t\) is time), then \(a(t)=\frac{kt^{2}}{m}\), and the velocity - time graph will be a curve (since \(v = v_0+\int_{0}^{t}a(t')dt'=v_0+\frac{k}{3m}t^{3}\), a cubic function, so the velocity - time graph is a curve with changing slope).
2. Displacement equation for constant acceleration: The displacement equation for constant acceleration is \(x = x_0+v_0t+\frac{1}{2}at^{2}\), which is derived under the assumption that \(a\) is constant. This equation comes from integrating the velocity function \(v = v_0 + at\) (which is linear in time for constant \(a\)) with respect to time.
3. Analyze non - constant acceleration case: For non - constant acceleration, \(a=a(t)\) (a function of time). The velocity function is \(v = v_0+\int_{0}^{t}a(t')dt'\), and the displacement function is \(x=x_0+\int_{0}^{t}v(t'')dt''=x_0+\int_{0}^{t}(v_0+\int_{0}^{t''}a(t')dt')dt''\). The standard constant - acceleration displacement equation assumes \(a\) is constant, so \(\int_{0}^{t}a(t')dt'=at\) and \(\int_{0}^{t}(v_0 + at'')dt''=v_0t+\frac{1}{2}at^{2}\). But when \(a\) is not constant, the integral of \(a(t)\) with respect to time is not a simple linear function of time (\(at\)), and thus the displacement equation \(x = x_0+v_0t+\frac{1}{2}at^{2}\) (derived for constant \(a\)) cannot account for the changing acceleration. For example, in the velocity - time graph with a curved line, the area under the curve (which represents displacement) cannot be calculated using the formula for the area of a trapezoid (which is what the constant - acceleration displacement formula is based on, as the velocity - time graph for constant acceleration is a straight line, and the area under it is the area of a trapezoid with bases \(v_0\) and \(v_0 + at\) and height \(t\)) but requires integration of the velocity function, which is more complex when \(a\) is not constant.

Answer:

1. Graph Example: A velocity - time graph of an object with non - constant acceleration (e.g., velocity as a cubic function of time, \(v(t)=v_0 + \frac{k}{3m}t^{3}\)) will be a curve (not a straight line). The slope of the velocity - time graph (acceleration) \(a(t)=\frac{dv}{dt}=\frac{kt^{2}}{m}\) changes with time.
2. Explanation of Displacement Equation Limitation: The displacement equation \(x = x_0+v_0t+\frac{1}{2}at^{2}\) is derived for constant acceleration (\(a\) is constant). For non - constant acceleration (\(a = a(t)\)), the acceleration changes with time. The velocity function is \(v=v_0+\int_{0}^{t}a(t')dt'\), and displacement is \(x = x_0+\int_{0}^{t}v(t'')dt''\). Since \(a(t)\) is not constant, \(\int_{0}^{t}a(t')dt'\) is not a linear function of time (\(at\)), so the simple quadratic displacement formula (for constant \(a\)) cannot accurately calculate displacement for non - constant acceleration. The area under the non - linear (curved) velocity - time graph (displacement) cannot be computed using the trapezoidal area formula (basis of the constant - acceleration displacement equation) and requires integration of the velocity function, which is more complex when \(a\) varies.