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 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 \(a<0\) since the slope is negative), then integrating \(v\) with respect to \(t\) gives \(x=x_0 + v_0t+\frac{1}{2}at^{2}\). This is a quadratic function of the form \(y = Ax^{2}+Bx + C\) (\(A=\frac{1}{2}a<0\), \(B = v_0>0\), \(C=x_0\)), which is a parabola opening downwards.

Step2: Check the position - time graphs

• Graph \(H\): The slope of the position - time graph (which represents velocity) is positive and decreasing (since the curve is getting flatter). But our velocity starts positive and decreases to zero.
• Graph \(I\): It is a straight - line graph. A straight - line position - time graph implies constant velocity (\(a = 0\)), which is not the case here.
• Graph \(J\): The slope of the position - time graph (velocity) is positive and decreasing. When \(v=\frac{dx}{dt}\), and \(v\) is a linear function of \(t\) (from a positive value to zero), \(x\) is a quadratic function. The general form of \(x\) (from \(v=v_0 + at\), \(a<0\)) is \(x=x_0+v_0t+\frac{1}{2}at^{2}\). The graph of \(x\) vs \(t\) for \(a<0\) is a parabola opening downwards. The slope of the position - time graph (velocity) starts positive and becomes zero.
• Graph \(K\): The slope of the position - time graph (velocity) is negative and increasing (in magnitude). But our initial velocity is positive.
• Graph \(L\): It is a straight - line graph. A straight - line position - time graph implies constant velocity (\(a = 0\)), which is not the case here.
• Graph \(M\): The slope of the position - time graph (velocity) is negative and decreasing (in magnitude). But our initial velocity is positive.

Answer:

\(J\)