Question

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

Explanation:

Step1: Recall the relationship between velocity and position

The velocity \(v\) is the derivative of the position \(x\) with respect to time \(t\), i.e., \(v=\frac{dx}{dt}\). If \(v = at + b\) (in this case, \(b = 0\) and \(a<0\)), then integrating \(v\) with respect to \(t\) gives \(x=\int vdt\). Since \(v=-kt\) (\(k>0\)), \(x=-\frac{1}{2}kt^{2}+C\) (where \(C\) is the initial position). The general form of the position - time function for \(v = mt\) (\(m<0\)) is a parabola. The slope of the position - time graph (which represents velocity) should be negative and its magnitude should be increasing with time.

Step2: Analyze each graph

• For a parabola \(x =-\frac{1}{2}kt^{2}+C\), the second - derivative \(\frac{d^{2}x}{dt^{2}}=-k<0\) (concave down).
• Graph \(R\) has a slope (velocity) that is negative and the magnitude of the slope (since the curve is getting steeper in the negative direction) is increasing with time.

Answer:

\(R\)