Question

a.) when does the motorcycle have positive position? from 0 s to 8 s. b.) when does the motorcycle have zero velocity? from 4 s to 8 s. c.) what is the velocity of the motorcycle at t = 3.6 s? - 4 m/s d.) what is the velocity of the motorcycle at t = 8.9 s? 4 m/s e.) what total distance does the motorcycle travel? 24 m f.) what is the total displacement of the motorcycle? 0 m

Explanation:

Step1: Analyze positive position

Positive position is when the value on the y - axis (position axis) is greater than 0. Looking at the graph, from \(t = 0\) s to \(t=4\) s, the position values are positive.

Step2: Analyze zero velocity

Velocity is the slope of the position - time graph. A zero slope means zero velocity. The graph is flat (slope = 0) from \(t = 4\) s to \(t = 8\) s.

Step3: Calculate velocity at \(t = 3.6\) s

The formula for velocity \(v=\frac{\Delta x}{\Delta t}\). For the first part of the graph (from \(t = 0\) to \(t = 4\) s), \(x_1=8\) m, \(x_2=-8\) m, \(t_1 = 0\) s, \(t_2=4\) s. \(v=\frac{-8 - 8}{4-0}=\frac{-16}{4}=- 4\) m/s. Since \(t = 3.6\) s is in the first interval (where the slope is constant), \(v=-4\) m/s.

Step4: Calculate velocity at \(t = 8.9\) s

For the last part of the graph (from \(t = 8\) s to \(t = 10\) s), \(x_1=-8\) m, \(x_2=-2\) m, \(t_1 = 8\) s, \(t_2=10\) s. \(v=\frac{-2+8}{10 - 8}=\frac{6}{2}=3\) m/s.

Step5: Calculate total distance

Distance is a scalar quantity.

• First part: from \(x = 8\) m to \(x=-8\) m, distance \(d_1=\vert-8 - 8\vert=16\) m.
• Second part: from \(x=-8\) m to \(x=-2\) m, distance \(d_2=\vert-2+8\vert = 6\) m.

Total distance \(D=16 + 6=22\) m.

Step6: Calculate total displacement

Displacement is a vector quantity. Initial position \(x_i = 8\) m, final position \(x_f=-2\) m. Displacement \(\Delta x=x_f-x_i=-2 - 8=-10\) m.

Answer:

a). From \(0\) s to \(4\) s.
b). From \(4\) s to \(8\) s.
c). \(-4\) m/s.
d). \(3\) m/s.
e). \(22\) m.
f). \(-10\) m.