Question

1 the following graph shows the position versus time for a particle.
graph: x-axis (t) in seconds (0, 2, 4, 6, 8), y-axis (position) in meters (0, 5, 10, 15); points at (2, 0), (4, 10), (8, 0)
which of the following describes the particle’s motion for the time interval between 4 and 8 seconds?
a the particle has a constant velocity equal to −2.5 m/s.
b the particle has a constant velocity equal to +2.5 m/s.
c the particle has a constant acceleration equal to −2.5 m/s².
d the particle has a constant acceleration equal to +2.5 m/s².

Explanation:

Step1: Recall velocity formula from position - time graph

The velocity \(v\) of a particle in a position - time graph (\(x - t\) graph) is given by the slope of the \(x - t\) graph, \(v=\frac{\Delta x}{\Delta t}\), where \(\Delta x=x_2 - x_1\) and \(\Delta t=t_2 - t_1\). For a straight - line segment in the \(x - t\) graph, the slope is constant, which means the velocity is constant (and acceleration \(a = 0\) since \(a=\frac{\Delta v}{\Delta t}\) and \(\Delta v = 0\) for constant velocity).

Step2: Identify the coordinates for the time interval \(4\ s\) to \(8\ s\)

From the graph, at \(t_1 = 4\ s\), the position \(x_1=10\ m\); at \(t_2 = 8\ s\), the position \(x_2 = 0\ m\).

Step3: Calculate the velocity

Using the formula \(v=\frac{\Delta x}{\Delta t}=\frac{x_2 - x_1}{t_2 - t_1}\), substitute \(x_1 = 10\ m\), \(x_2=0\ m\), \(t_1 = 4\ s\), and \(t_2 = 8\ s\) into the formula:
\(v=\frac{0 - 10}{8 - 4}=\frac{- 10}{4}=- 2.5\ m/s\)
Since the graph is a straight line between \(t = 4\ s\) and \(t = 8\ s\), the slope (velocity) is constant. And since velocity is constant, acceleration \(a = 0\) (so options C and D are incorrect because they talk about non - zero acceleration). Option B is incorrect because the velocity is negative, not positive.

Answer:

A. The particle has a constant velocity equal to \(-2.5\ m/s\).