Question

velocity-time graph
what is the objects instantaneous velocity at t = 25 s?
0 m/s
4 m/s
10 m/s
25 m/s

Explanation:

Step1: Analyze the Velocity - Time Graph

In a velocity - time graph, the velocity at a particular time \(t\) is given by the \(y\) - coordinate (velocity axis) corresponding to the \(x\) - coordinate (time axis) of \(t\). We need to find the velocity at \(t = 25\ s\).

Looking at the graph, the time axis (\(x\) - axis) has values like \(10\ s\), \(20\ s\), \(30\ s\) etc. The velocity axis (\(y\) - axis) has values for velocity. At \(t=25\ s\), we look at the point on the graph corresponding to \(t = 25\ s\). From the graph, we can see that the velocity at \(t = 25\ s\) is \(0\ m/s\) (since the graph crosses the time axis (where velocity \(v = 0\)) at around \(t=25\ s\)).

Wait, there seems to be a mistake in the initial thought. Let's re - examine. Wait, the graph: from \(t = 20\ s\) to \(t\) where it crosses the \(t\) - axis (velocity = 0), the velocity is decreasing from \(10\ m/s\) to \(0\ m/s\)? Wait no, actually, looking at the graph, the line from \(t = 20\ s\) (where velocity is \(10\ m/s\)) to \(t\) (let's say \(t = 25\ s\) is on that line). Wait, no, the graph: the first part is increasing from \(t = 0\) to \(t = 10\ s\) to \(10\ m/s\), then constant from \(t = 10\ s\) to \(t = 20\ s\) at \(10\ m/s\), then decreasing from \(t = 20\ s\) to \(t\) (let's find when it crosses \(v = 0\)). Wait, the time when \(v = 0\) is at \(t = 25\ s\)? Wait, no, let's check the options. The options are \(0\ m/s\), \(4\ m/s\), \(10\ m/s\), \(25\ m/s\).

Wait, maybe the graph is such that at \(t = 25\ s\), the velocity is \(0\ m/s\) because the graph crosses the \(v = 0\) line at \(t = 25\ s\).

Step2: Confirm the Velocity at \(t = 25\ s\)

From the velocity - time graph, the point corresponding to \(t=25\ s\) lies on the line where velocity is \(0\ m/s\) (the line that crosses the time - axis, i.e., \(v = 0\) line). So the instantaneous velocity at \(t = 25\ s\) is \(0\ m/s\).

Answer:

\(0\ m/s\)