Question

1. use the velocity - time graph 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
if the area under the line forms a ... triangle, then use area = 0.5 baseheight
if the area under the line forms a ... 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.

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is change in time.

Step2: Analyze first graph

For the first velocity - time graph (left - hand side of problem 3), initial velocity $v_1 = 0\ m/s$, final velocity $v_2=20\ m/s$, and time $\Delta t = 8\ s$. Then $a_1=\frac{v_2 - v_1}{\Delta t}=\frac{20 - 0}{8}=2.5\ m/s^{2}$.

Step3: Analyze second graph

For the second velocity - time graph (right - hand side of problem 3), initial velocity $v_1 = 30\ m/s$, final velocity $v_2 = 0\ m/s$, and time $\Delta t=12\ s$. Then $a_2=\frac{v_2 - v_1}{\Delta t}=\frac{0 - 30}{12}=- 2.5\ m/s^{2}$.

For problem 4:

Step1: Analyze first velocity - time graph

The first velocity - time graph in problem 4 is a rectangle. Area (displacement) $A_1=v\times t$, where $v = 8\ m/s$ and $t = 8\ s$. So $A_1=8\times8 = 64\ m$.

Step2: Analyze second velocity - time graph

The second velocity - time graph in problem 4 is a triangle. Area (displacement) $A_2=\frac{1}{2}\times v\times t$, where $v = 12\ m/s$ and $t = 8\ s$. So $A_2=\frac{1}{2}\times12\times8=48\ m$.

Step3: Analyze third velocity - time graph

The third velocity - time graph in problem 4 is a trapezoid. We can split it into a rectangle and a triangle. The rectangle has $v_1 = 4\ m/s$ and $t = 8\ s$, and the triangle has $v_2=12 - 4=8\ m/s$ and $t = 8\ s$. Area (displacement) $A_3=(4\times8)+\frac{1}{2}\times8\times8=32 + 32=64\ m$.

Answer:

For problem 3, the accelerations are $2.5\ m/s^{2}$ and $-2.5\ m/s^{2}$. For problem 4, the displacements are $64\ m$, $48\ m$, and $64\ m$.