Question

1. use the velocity - time graphs below to determine the acceleration. psyw
2. the area under the line of a velocity - time graph can be calculated using simple rectangle and triangle equations. the graphs below are examples:

if the area under the line forms a ...
... rectangle, then use area = base*height
... triangle, then use area = 0.5 baseheight
... trapezoid, then make it into a rectangle + triangle and add the two areas.
a=(6 m/s)*(6 s)=36 m
a = 0.5(6 m/s)(6 s)=18 m
atotal = a rectangle+atriangle
atotal=(2 m/s)(6 s)+0.5(4 m/s)*(6 s)=24 m
find the displacement of the objects represented by the following velocity - time graphs.

1. for the following pos - time graphs, determine the corresponding shape of the vel - time graph.

Explanation:

1. For the problem of finding acceleration from velocity - time graphs (Question 3):

• Explanation:
• Step 1: Recall the formula for acceleration
• Acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) (final velocity - initial velocity) and \(\Delta t=t_f - t_i\) (final time - initial time).
• For the first velocity - time graph in Question 3:
• \(v_i = 5\ m/s\), \(v_f=30\ m/s\), \(t_i = 0\ s\), \(t_f = 8\ s\).
• \(a=\frac{v_f - v_i}{t_f - t_i}=\frac{30 - 5}{8-0}=\frac{25}{8}=3.125\ m/s^{2}\).
• For the second velocity - time graph in Question 3:
• \(v_i = 30\ m/s\), \(v_f = 5\ m/s\), \(t_i=0\ s\), \(t_f = 12\ s\).
• \(a=\frac{v_f - v_i}{t_f - t_i}=\frac{5 - 30}{12 - 0}=\frac{- 25}{12}\approx - 2.083\ m/s^{2}\).
• Answer: For the first graph, the acceleration is \(3.125\ m/s^{2}\). For the second graph, the acceleration is \(-2.083\ m/s^{2}\).

1. For the problem of finding displacement from velocity - time graphs (Question 4):

• Explanation:
• For the first velocity - time graph in Question 4:
• The graph is a rectangle with \(v = 12\ m/s\) and \(t = 8\ s\). Using the formula for the area of a rectangle \(A=v\times t\).
• \(A=(12\ m/s)\times(8\ s)=96\ m\).
• For the second velocity - time graph in Question 4:
• The graph is a triangle with \(v_i = 0\ m/s\), \(v_f = 12\ m/s\) and \(t = 8\ s\). Using the formula for the area of a triangle \(A = 0.5\times v\times t\).
• \(A=0.5\times(12\ m/s)\times(8\ s)=48\ m\).
• For the third velocity - time graph in Question 4:
• The graph can be split into a rectangle and a triangle. The rectangle has \(v_{rect}=4\ m/s\) and \(t = 8\ s\), and the triangle has \(v_{triangle}=12 - 4=8\ m/s\) and \(t = 8\ s\).
• \(A_{rect}=(4\ m/s)\times(8\ s)=32\ m\), \(A_{triangle}=0.5\times(8\ m/s)\times(8\ s)=32\ m\).
• \(A_{total}=A_{rect}+A_{triangle}=32 + 32=64\ m\).
• Answer: For the first graph, the displacement is \(96\ m\). For the second graph, the displacement is \(48\ m\). For the third graph, the displacement is \(64\ m\).

Answer:

1. For the problem of finding acceleration from velocity - time graphs (Question 3):

• Explanation:
• Step 1: Recall the formula for acceleration
• Acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) (final velocity - initial velocity) and \(\Delta t=t_f - t_i\) (final time - initial time).
• For the first velocity - time graph in Question 3:
• \(v_i = 5\ m/s\), \(v_f=30\ m/s\), \(t_i = 0\ s\), \(t_f = 8\ s\).
• \(a=\frac{v_f - v_i}{t_f - t_i}=\frac{30 - 5}{8-0}=\frac{25}{8}=3.125\ m/s^{2}\).
• For the second velocity - time graph in Question 3:
• \(v_i = 30\ m/s\), \(v_f = 5\ m/s\), \(t_i=0\ s\), \(t_f = 12\ s\).
• \(a=\frac{v_f - v_i}{t_f - t_i}=\frac{5 - 30}{12 - 0}=\frac{- 25}{12}\approx - 2.083\ m/s^{2}\).
• Answer: For the first graph, the acceleration is \(3.125\ m/s^{2}\). For the second graph, the acceleration is \(-2.083\ m/s^{2}\).

1. For the problem of finding displacement from velocity - time graphs (Question 4):

• Explanation:
• For the first velocity - time graph in Question 4:
• The graph is a rectangle with \(v = 12\ m/s\) and \(t = 8\ s\). Using the formula for the area of a rectangle \(A=v\times t\).
• \(A=(12\ m/s)\times(8\ s)=96\ m\).
• For the second velocity - time graph in Question 4:
• The graph is a triangle with \(v_i = 0\ m/s\), \(v_f = 12\ m/s\) and \(t = 8\ s\). Using the formula for the area of a triangle \(A = 0.5\times v\times t\).
• \(A=0.5\times(12\ m/s)\times(8\ s)=48\ m\).
• For the third velocity - time graph in Question 4:
• The graph can be split into a rectangle and a triangle. The rectangle has \(v_{rect}=4\ m/s\) and \(t = 8\ s\), and the triangle has \(v_{triangle}=12 - 4=8\ m/s\) and \(t = 8\ s\).
• \(A_{rect}=(4\ m/s)\times(8\ s)=32\ m\), \(A_{triangle}=0.5\times(8\ m/s)\times(8\ s)=32\ m\).
• \(A_{total}=A_{rect}+A_{triangle}=32 + 32=64\ m\).
• Answer: For the first graph, the displacement is \(96\ m\). For the second graph, the displacement is \(48\ m\). For the third graph, the displacement is \(64\ m\).