Question

14 the following graph shows the velocity versus time for a particle moving in front of a sonic motion detector. which of the following describes the particle’s acceleration for the time interval between 0 and 3 seconds?

Explanation:

To determine the acceleration, we analyze the velocity - time graph in three intervals: \(0 - 1\) s, \(1 - 2\) s, and \(2 - 3\) s.

Step 1: Analyze the interval \(0 - 1\) s

Acceleration \(a=\frac{\Delta v}{\Delta t}\). In the time interval from \(t = 0\) to \(t=1\) s, the initial velocity \(v_i = 10\) m/s and the final velocity \(v_f=5\) m/s, and \(\Delta t=1 - 0 = 1\) s.
Using the formula for acceleration \(a=\frac{v_f - v_i}{\Delta t}\), we substitute the values: \(a=\frac{5 - 10}{1-0}=\frac{- 5}{1}=- 5\) m/s². The negative sign indicates that the acceleration is constant and directed opposite to the velocity (deceleration) in this interval.

Step 2: Analyze the interval \(1 - 2\) s

In the time interval from \(t = 1\) to \(t = 2\) s, the velocity remains constant (\(v = 5\) m/s). The change in velocity \(\Delta v=5 - 5=0\) m/s and \(\Delta t=2 - 1 = 1\) s.
Using the formula \(a=\frac{\Delta v}{\Delta t}\), we get \(a=\frac{0}{1}=0\) m/s². So, the acceleration is zero in this interval.

Step 3: Analyze the interval \(2 - 3\) s

In the time interval from \(t = 2\) to \(t=3\) s, the initial velocity \(v_i = 5\) m/s and the final velocity \(v_f = 10\) m/s, and \(\Delta t=3 - 2=1\) s.
Using the formula for acceleration \(a=\frac{v_f - v_i}{\Delta t}\), we substitute the values: \(a=\frac{10 - 5}{3 - 2}=\frac{5}{1} = 5\) m/s². The acceleration is constant and positive (speeding up in the direction of motion) in this interval.

So, the acceleration of the particle is \(- 5\) m/s² from \(0\) to \(1\) s, \(0\) m/s² from \(1\) to \(2\) s, and \(5\) m/s² from \(2\) to \(3\) s. In other words, the particle has a constant negative acceleration (deceleration) from \(0\) to \(1\) second, zero acceleration from \(1\) to \(2\) seconds, and a constant positive acceleration (acceleration) from \(2\) to \(3\) seconds.

If we were to describe it in terms of the overall behavior:

• From \(t = 0\) to \(t = 1\) s: The acceleration is a constant \(- 5\) m/s² (the slope of the velocity - time graph is negative and constant).
• From \(t=1\) to \(t = 2\) s: The acceleration is \(0\) m/s² (the slope of the velocity - time graph is zero as the velocity is constant).
• From \(t = 2\) to \(t=3\) s: The acceleration is a constant \(5\) m/s² (the slope of the velocity - time graph is positive and constant).

If we assume that the answer is expected in terms of the three - part behavior:

Answer:

From \(0\) to \(1\) s, acceleration \(a=- 5\) m/s² (constant deceleration); from \(1\) to \(2\) s, \(a = 0\) m/s² (constant velocity); from \(2\) to \(3\) s, \(a=5\) m/s² (constant acceleration)