Question

8 a dune buggy is cruising along a beach. its motion is recorded in the velocity versus time graph below. what is the total distance the dune buggy traveled from 0 to 12 seconds? f 100 m g 350 m h 440 m j 540 m

Explanation:

To find the total distance traveled, we calculate the area under the velocity - time graph from \(t = 0\) to \(t=12\) seconds. The graph can be divided into geometric shapes (triangles, rectangles, trapezoids), and we find the area of each shape and sum them up.

Step 1: Analyze the first segment (\(t = 0\) to \(t = 4\) seconds)

This is a trapezoid (or a triangle and a rectangle, but we can use the trapezoid area formula \(A=\frac{(v_1 + v_2)}{2}\times t\), where \(v_1\) and \(v_2\) are the initial and final velocities, and \(t\) is the time interval). At \(t = 0\), \(v_1=35\space m/s\) (from the graph, the initial velocity is between 30 and 40, looking at the grid, each grid line is 5 m/s? Wait, actually, let's re - examine the graph. From \(t = 0\) to \(t = 4\) seconds, the velocity starts at \(v_0 = 35\space m/s\) (wait, no, looking at the y - axis: the first point at \(t = 0\) is at \(y = 35\) (since between 30 and 40, mid - way? Wait, actually, the graph from \(t = 0\) to \(t = 4\) seconds: at \(t = 0\), \(v=35\space m/s\), at \(t = 4\) seconds, \(v = 15\space m/s\), and the time interval \(\Delta t_1=4 - 0=4\space s\). The area of a trapezoid is \(A_1=\frac{(v_0 + v_4)}{2}\times\Delta t_1=\frac{(35 + 15)}{2}\times4=\frac{50}{2}\times4 = 100\space m\)

Step 2: Analyze the second segment (\(t = 4\) to \(t = 6\) seconds)

This is a rectangle with velocity \(v = 15\space m/s\) and time interval \(\Delta t_2=6 - 4 = 2\space s\). The area of a rectangle is \(A_2=v\times\Delta t_2=15\times2 = 30\space m\)

Step 3: Analyze the third segment (\(t = 6\) to \(t = 8\) seconds)

This is a triangle (or a trapezoid) with initial velocity \(v_6 = 15\space m/s\), final velocity \(v_8=45\space m/s\), and time interval \(\Delta t_3=8 - 6 = 2\space s\). Using the trapezoid area formula \(A_3=\frac{(v_6 + v_8)}{2}\times\Delta t_3=\frac{(15 + 45)}{2}\times2=\frac{60}{2}\times2=60\space m\)

Step 4: Analyze the fourth segment (\(t = 8\) to \(t = 10\) seconds)

This is a rectangle with velocity \(v = 45\space m/s\) and time interval \(\Delta t_4=10 - 8=2\space s\). The area of the rectangle is \(A_4 = 45\times2=90\space m\)

Step 5: Analyze the fifth segment (\(t = 10\) to \(t = 12\) seconds)

This is a trapezoid with initial velocity \(v_{10}=45\space m/s\), final velocity \(v_{12}=25\space m/s\), and time interval \(\Delta t_5=12 - 10 = 2\space s\). Using the trapezoid area formula \(A_5=\frac{(v_{10}+v_{12})}{2}\times\Delta t_5=\frac{(45 + 25)}{2}\times2=\frac{70}{2}\times2 = 70\space m\)

Wait, maybe a better way is to divide the graph into parts:

1. From \(t = 0\) to \(t = 4\): The velocity - time graph is a line from \((0,35)\) to \((4,15)\). The area under this line is the area of a trapezoid with bases \(b_1 = 35\), \(b_2=15\) and height \(h = 4\). \(A_1=\frac{(35 + 15)\times4}{2}=100\)

2. From \(t = 4\) to \(t = 6\): The velocity is constant at \(15\space m/s\), so it's a rectangle with length \(l = 2\) and width \(w = 15\). \(A_2=15\times2 = 30\)

3. From \(t = 6\) to \(t = 8\): The velocity goes from \(15\) to \(45\), so it's a trapezoid with bases \(15\) and \(45\) and height \(2\). \(A_3=\frac{(15 + 45)\times2}{2}=60\)

4. From \(t = 8\) to \(t = 10\): The velocity is constant at \(45\space m/s\), so it's a rectangle with length \(2\) and width \(45\). \(A_4=45\times2=90\)

5. From \(t = 10\) to \(t = 12\): The velocity goes from \(45\) to \(25\), so it's a trapezoid with bases \(45\) and \(25\) and height \(2\). \(A_5=\frac{(45 + 25)\times2}{2}=70\)

Now, sum up all the areas: \(A=A_1+A_2+A_3+A_4+A_5=100 + 30+60 + 90+70=350\space m\)

Wait, maybe my initial velocity at \(t = 0\) was wrong. Let's re - examine the graph. The y - axis: each major grid line is 10 m/s? Wait, the first point at \(t = 0\) is at \(y = 35\)? No, looking at the graph, at \(t = 0\), the velocity is \(35\) (between 30 and 40, with each small grid being 5 m/s). At \(t = 4\), it's \(15\) (between 10 and 20). At \(t = 6\), it's \(15\). At \(t = 8\), it's \(45\) (between 40 and 50). At \(t = 10\), it's \(45\). At \(t = 12\), it's \(25\) (between 20 and 30).

Another way: The area under a velocity - time graph is distance. We can also divide the graph into:

- From \(t = 0\) to \(t = 4\): A trapezoid with \(v_1 = 35\), \(v_2=15\), \(t = 4\). Area \(=\frac{(35 + 15)\times4}{2}=100\)

- From \(t = 4\) to \(t = 6\): A rectangle with \(v = 15\), \(t = 2\). Area \(=15\times2 = 30\)

- From \(t = 6\) to \(t = 8\): A triangle? No, it's a trapezoid with \(v_1 = 15\), \(v_2 = 45\), \(t = 2\). Area \(=\frac{(15 + 45)\times2}{2}=60\)

- From \(t = 8\) to \(t = 10\): A rectangle with \(v = 45\), \(t = 2\). Area \(=45\times2=90\)

- From \(t = 10\) to \(t = 12\): A trapezoid with \(v_1 = 45\), \(v_2 = 25\), \(t = 2\). Area \(=\frac{(45 + 25)\times2}{2}=70\)

Summing these: \(100+30 + 60+90+70=350\)

Answer:

G 350 m