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:

... rectangle, then use area = base*height
if the area under the line forms a ... 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=(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

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 velocity - time graph

For the first velocity - time graph (left - hand side of top part), initial velocity $v_1 = 0\ m/s$, final velocity $v_2=20\ m/s$, time $t_1 = 0\ s$, $t_2 = 8\ s$. Then $\Delta v=v_2 - v_1=20 - 0=20\ m/s$ and $\Delta t=t_2 - t_1=8 - 0 = 8\ s$. So $a=\frac{\Delta v}{\Delta t}=\frac{20}{8}=2.5\ m/s^{2}$.

Step3: Analyze second velocity - time graph

For the second velocity - time graph (right - hand side of top part), initial velocity $v_1 = 30\ m/s$, final velocity $v_2 = 0\ m/s$, time $t_1 = 0\ s$, $t_2=12\ s$. Then $\Delta v=v_2 - v_1=0 - 30=- 30\ m/s$ and $\Delta t=t_2 - t_1=12 - 0 = 12\ s$. So $a=\frac{\Delta v}{\Delta t}=\frac{-30}{12}=-2.5\ m/s^{2}$.

Step4: Recall displacement formula for velocity - time graph

Displacement $d$ is the area under the velocity - time graph.

Step5: Calculate displacement for first bottom graph

For the first bottom graph, it is a rectangle. Area $A = v\times t$, where $v = 12\ m/s$ and $t = 8\ s$. So $d=12\times8 = 96\ m$.

Step6: Calculate displacement for second bottom graph

For the second bottom graph, it is a triangle. Area $A=\frac{1}{2}\times base\times height$, where base $t = 8\ s$ and height $v = 12\ m/s$. So $d=\frac{1}{2}\times8\times12=48\ m$.

Step7: Calculate displacement for third bottom graph

For the third bottom graph, it is a trapezoid. We can split it into a rectangle and a triangle. The rectangle has height $v_1 = 4\ m/s$ and base $t = 8\ s$, and the triangle has base $t = 8\ s$ and height $v_2=12 - 4 = 8\ m/s$. The area of the rectangle $A_1=4\times8 = 32\ m$, the area of the triangle $A_2=\frac{1}{2}\times8\times8=32\ m$. So $d=A_1 + A_2=32+32 = 64\ m$.

Answer:

1. Acceleration of first top graph: $2.5\ m/s^{2}$
2. Acceleration of second top graph: $-2.5\ m/s^{2}$
3. Displacement of first bottom graph: $96\ m$
4. Displacement of second bottom graph: $48\ m$
5. Displacement of third bottom graph: $64\ m$