Question

1. given the velocity(m/s) vs. time(s) graph below, answer each of the following. v(t)(meters per second) (0, 4) (8, 0) t(seconds) a. find the acceleration of the particle from t = 0 to t = 8. show how you got your answer. \\(\frac{0-4}{8-0}=\frac{-4}{8}=-\frac{1}{2}\\) b. what is the net change in displacement of the particle from t = 0 to t = 8? show how you got your answer. c. what is the velocity of the particle at t = 4? use the graph to answer

Explanation:

Part a

Step1: Recall acceleration formula

Acceleration \(a\) is the slope of the velocity - time graph, given by \(a=\frac{\Delta v}{\Delta t}=\frac{v(t_2)-v(t_1)}{t_2 - t_1}\)
We have \(t_1 = 0\), \(v(t_1)=4\ m/s\), \(t_2 = 8\), \(v(t_2)=0\ m/s\)

Step2: Calculate acceleration

Substitute the values into the formula: \(a=\frac{0 - 4}{8-0}=\frac{- 4}{8}=-\frac{1}{2}\ m/s^{2}\)

Step1: Recall displacement formula

The net change in displacement for a velocity - time graph is the area under the velocity - time curve. The graph is a triangle with base \(b = 8\ s\) and height \(h=4\ m/s\)
The formula for the area of a triangle is \(A=\frac{1}{2}\times base\times height\)

Step2: Calculate displacement

Substitute \(base = 8\) and \(height = 4\) into the formula: \(A=\frac{1}{2}\times8\times4=16\ meters\)

Step1: Analyze the velocity - time graph

The velocity - time graph is a straight line (linear) from \((0,4)\) to \((8,0)\). The equation of the line can be found using the slope - intercept form \(y=mx + c\), where \(m\) is the slope and \(c\) is the y - intercept. We know \(m=-\frac{1}{2}\) (from part a) and \(c = 4\), so the equation of the velocity function is \(v(t)=-\frac{1}{2}t + 4\)

Step2: Find velocity at \(t = 4\)

Substitute \(t = 4\) into \(v(t)=-\frac{1}{2}t + 4\): \(v(4)=-\frac{1}{2}(4)+4=- 2 + 4=2\ m/s\)
Alternatively, since the graph is a straight line, at \(t = 4\) (which is the mid - point of \(t = 0\) and \(t = 8\)), the velocity is the average of the initial and final velocities. The initial velocity \(v(0)=4\ m/s\) and final velocity \(v(8) = 0\ m/s\), so \(v(4)=\frac{4 + 0}{2}=2\ m/s\)

Answer:

The acceleration is \(-\frac{1}{2}\ m/s^{2}\)

Part b