Question

1. the graph above shows the velocity versus time for an object moving in a straight line. at what time after t = 0 does the object again pass through its initial position? (a) between 0 and 1 s (b) 1 s (c) between 1 and 2 s (d) 2s

Explanation:

Step1: Recall the relationship between velocity - time graph and displacement

The displacement of an object is given by the area under the velocity - time graph. When the net area (taking the area above the time - axis as positive and the area below the time - axis as negative) is zero, the object is at its initial position.

Step2: Analyze the area under the graph

For \(t\in[0,1]\) s: The area below the \(t\) - axis (negative displacement) and for \(t\in[1,2]\) s: The area above the \(t\) - axis (positive displacement).
Let's assume the area of the triangle from \(t = 0\) to \(t=1\) is \(A_1\) (negative) and the area of the trapezoid (or part of the graph) from \(t = 1\) to some \(t\in(1,2)\) is \(A_2\) (positive).
The area of a triangle is \(A=\frac{1}{2}\times base\times height\). The triangle from \(t = 0\) to \(t = 1\) has a base \(b = 1\) s and height \(h=- 1\) m/s (using the symmetry of the first - part of the graph), so \(|A_1|=\frac{1}{2}\times1\times1 = 0.5\) \(m\).
As we start adding the positive area (from \(t = 1\) s) which is increasing, at a time between \(t = 1\) s and \(t=2\) s, the sum of the negative area (from \(t = 0\) to \(t = 1\) s) and the positive area (from \(t = 1\) s to some \(t\in(1,2)\) s) will be zero.

Answer:

C. Between 1 and 2 s