Question

motion in one dimension

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 = 0.5baseheight
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+a triangle
atotal=(2m/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.

© the physics classroom, 2009
page 16

Explanation:

1. For question 3 (finding acceleration from velocity - time graph):

• Left - hand graph:
• # Explanation:

Step1: Recall acceleration formula

The acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t=t_f - t_i\). From the graph, \(v_i = 5\ m/s\), \(v_f=30\ m/s\), \(t_i = 0\ s\), \(t_f = 8\ s\).
\(\Delta v=30 - 5=25\ m/s\), \(\Delta t = 8-0 = 8\ s\).
\(a=\frac{30 - 5}{8}=\frac{25}{8}=3.125\ m/s^{2}\)

Step2: No further steps needed

• # Answer: \(3.125\ m/s^{2}\)
• Right - hand graph:
• # Explanation:

Step1: Recall acceleration formula

The acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t=t_f - t_i\). From the graph, \(v_i = 30\ m/s\), \(v_f = 10\ m/s\), \(t_i=0\ s\), \(t_f = 12\ s\).
\(\Delta v=10 - 30=- 20\ m/s\), \(\Delta t=12 - 0 = 12\ s\).
\(a=\frac{10 - 30}{12}=-\frac{20}{12}=-\frac{5}{3}\approx - 1.67\ m/s^{2}\)

Step2: No further steps needed

• # Answer: \(-\frac{5}{3}\ m/s^{2}\approx - 1.67\ m/s^{2}\)

1. For question 4 (finding displacement from velocity - time graph):

• Left - hand graph:
• # Explanation:

Step1: Identify shape and formula

The graph is a rectangle. The formula for the area of a rectangle is \(A = base\times height\). Here, the base (time) \(t = 8\ s\) and the height (velocity) \(v = 8\ m/s\).
\(A=(8\ m/s)\times(8\ s)=64\ m\)

Step2: No further steps needed

• # Answer: \(64\ m\)
• Middle graph:
• # Explanation:

Step1: Identify shape and formula

The graph is a triangle. The formula for the area of a triangle is \(A = 0.5\times base\times height\). Here, the base (time) \(t = 8\ s\) and the height (velocity) \(v = 12\ m/s\).
\(A = 0.5\times(8\ s)\times(12\ m/s)=48\ m\)

Step2: No further steps needed

• # Answer: \(48\ m\)
• Right - hand graph:
• # Explanation:

Step1: Split into rectangle and triangle

Split the shape into a rectangle and a triangle. For the rectangle, base \(t_1=4\ s\), height \(v_1 = 4\ m/s\), so \(A_{rectangle}=(4\ m/s)\times(4\ s)=16\ m\). For the triangle, base \(t_2 = 8\ s\), height \(v_2=12 - 4=8\ m/s\), so \(A_{triangle}=0.5\times(8\ s)\times(8\ m/s)=32\ m\).

Step2: Calculate total area

\(A_{total}=A_{rectangle}+A_{triangle}=16 + 32=48\ m\)

• # Answer: \(48\ m\)

Answer:

1. For question 3 (finding acceleration from velocity - time graph):

• Left - hand graph:
• # Explanation:

Step1: Recall acceleration formula

The acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t=t_f - t_i\). From the graph, \(v_i = 5\ m/s\), \(v_f=30\ m/s\), \(t_i = 0\ s\), \(t_f = 8\ s\).
\(\Delta v=30 - 5=25\ m/s\), \(\Delta t = 8-0 = 8\ s\).
\(a=\frac{30 - 5}{8}=\frac{25}{8}=3.125\ m/s^{2}\)

Step2: No further steps needed

• # Answer: \(3.125\ m/s^{2}\)
• Right - hand graph:
• # Explanation:

Step1: Recall acceleration formula

The acceleration \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v = v_f - v_i\) and \(\Delta t=t_f - t_i\). From the graph, \(v_i = 30\ m/s\), \(v_f = 10\ m/s\), \(t_i=0\ s\), \(t_f = 12\ s\).
\(\Delta v=10 - 30=- 20\ m/s\), \(\Delta t=12 - 0 = 12\ s\).
\(a=\frac{10 - 30}{12}=-\frac{20}{12}=-\frac{5}{3}\approx - 1.67\ m/s^{2}\)

Step2: No further steps needed

• # Answer: \(-\frac{5}{3}\ m/s^{2}\approx - 1.67\ m/s^{2}\)

1. For question 4 (finding displacement from velocity - time graph):

• Left - hand graph:
• # Explanation:

Step1: Identify shape and formula

The graph is a rectangle. The formula for the area of a rectangle is \(A = base\times height\). Here, the base (time) \(t = 8\ s\) and the height (velocity) \(v = 8\ m/s\).
\(A=(8\ m/s)\times(8\ s)=64\ m\)

Step2: No further steps needed

• # Answer: \(64\ m\)
• Middle graph:
• # Explanation:

Step1: Identify shape and formula

The graph is a triangle. The formula for the area of a triangle is \(A = 0.5\times base\times height\). Here, the base (time) \(t = 8\ s\) and the height (velocity) \(v = 12\ m/s\).
\(A = 0.5\times(8\ s)\times(12\ m/s)=48\ m\)

Step2: No further steps needed

• # Answer: \(48\ m\)
• Right - hand graph:
• # Explanation:

Step1: Split into rectangle and triangle

Split the shape into a rectangle and a triangle. For the rectangle, base \(t_1=4\ s\), height \(v_1 = 4\ m/s\), so \(A_{rectangle}=(4\ m/s)\times(4\ s)=16\ m\). For the triangle, base \(t_2 = 8\ s\), height \(v_2=12 - 4=8\ m/s\), so \(A_{triangle}=0.5\times(8\ s)\times(8\ m/s)=32\ m\).

Step2: Calculate total area

\(A_{total}=A_{rectangle}+A_{triangle}=16 + 32=48\ m\)

• # Answer: \(48\ m\)