Question

the accompanying figure shows the velocity v = ds/dt = f(t) (m/sec) of a body moving along a coordinate line. a. when does the body reverse direction? b. when is it moving at a constant speed? c. graph the body’s speed for 0 ≤ t ≤ 10. d. graph the acceleration, where defined.

Explanation:

Step1: Determine direction - change points

The body reverses direction when velocity changes sign. From the graph, this occurs at \(t = 2\) and \(t=4\) seconds.

Step2: Identify constant - speed intervals

Constant speed means \(|v|\) is constant. The body moves at a constant speed when \(5\leq t\leq8\) (since \(v = 5\) m/s in this interval).

Step3: Graph speed

Speed is \(|v|\). For \(0\leq t\leq2\), \(v\) is increasing from \(0\) to \(5\), so speed is also increasing from \(0\) to \(5\). At \(t = 2\), speed starts decreasing to \(0\) at \(t = 3\), then increases to \(5\) at \(t = 4\). From \(t=5\) to \(t = 8\), speed is \(5\), and then it decreases to \(0\) at \(t = 10\).

Step4: Graph acceleration

Acceleration \(a=\frac{dv}{dt}\). For \(0\lt t\lt2\), \(a=\frac{5 - 0}{2-0}=2.5\) m/s². At \(t = 2\), \(a\) is not - defined (discontinuity). For \(2\lt t\lt3\), \(a=\frac{- 5-5}{3 - 2}=-10\) m/s². At \(t = 3\), \(a\) is not - defined. For \(3\lt t\lt4\), \(a=\frac{5+5}{4 - 3}=10\) m/s². At \(t = 4\), \(a\) is not - defined. For \(5\lt t\lt8\), \(a = 0\). For \(8\lt t\lt10\), \(a=\frac{0 - 5}{10 - 8}=-2.5\) m/s².

Answer:

a. \(t = 2,4\)
b. \(5\leq t\leq8\)
c. Graph speed by plotting \(|v|\) values as described above for \(0\leq t\leq10\).
d. Graph acceleration by plotting the values calculated above for the intervals where it is defined.