Question

which position - time graph is equivalent to the given velocity - time graph?

Explanation:

Step1: Analyze the velocity - time graph

The velocity - time graph has a negative slope (velocity is decreasing linearly from a positive value to zero). The area under the velocity - time graph gives the displacement. The formula for velocity \(v=\frac{dx}{dt}\), and if \(v = v_0+at\) (where \(v_0> 0\) and \(a<0\)), then integrating \(v\) with respect to \(t\) to get \(x\): \(x=\int vdt=\int(v_0 + at)dt=v_0t+\frac{1}{2}at^{2}+x_0\). This is a quadratic function \(y = Ax^{2}+Bx + C\) (\(A=\frac{a}{2}<0\), \(B = v_0>0\), \(C=x_0\)), and the graph of a quadratic function \(y = Ax^{2}+Bx + C\) (\(A<0\)) is a parabola opening downwards.

Step2: Check the position - time graphs

• Graph \(I\): Is a straight - line graph. For a straight - line position - time graph, \(v=\frac{dx}{dt}\) is constant (since the slope of \(x - t\) graph is velocity). But our velocity is changing, so \(I\) is incorrect.
• Graph \(J\): Is a parabola opening downwards. Since \(v = v_0+at\) (\(v_0>0\), \(a < 0\)), \(x=v_0t+\frac{1}{2}at^{2}+x_0\) (a quadratic with \(x(0)=x_0\) and the coefficient of \(t^{2}\) negative).
• Graph \(K\): Is a parabola opening upwards. For \(x = v_0t+\frac{1}{2}at^{2}+x_0\) with \(a<0\), the parabola should open downwards, so \(K\) is incorrect.
• Graphs \(H, L, M\): Do not represent the quadratic function \(x=v_0t+\frac{1}{2}at^{2}+x_0\) ( \(H\) has a non - zero initial slope but the curvature is not consistent with \(a<0\) for all time, \(L\) has a negative initial velocity assumption which is wrong as \(v(0)>0\), \(M\) has a negative initial velocity assumption which is wrong as \(v(0)>0\))

Answer:

\(J\)