Question

the accompanying figure shows the velocity v = \frac{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 bodys 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, velocity changes from positive to negative or vice - versa at \(t = 5\) sec.

Step2: Identify constant - speed intervals

Constant speed means \(|v|\) is constant. From the graph, \(|v|\) is constant on the intervals \([1,4]\) and \([7,10]\).

Step3: Graph speed

Speed is \(|v|\). For \(0\leq t<1\), \(v\) is increasing from \(0\) to a positive value; for \(1\leq t\leq4\), \(v\) is a positive constant; for \(4 < t<5\), \(v\) is decreasing from the positive constant to \(0\); for \(5 < t<7\), \(v\) is negative and increasing in magnitude; for \(7\leq t\leq10\), \(v\) is a negative constant.

Step4: Graph acceleration

Acceleration \(a=\frac{dv}{dt}\). It is the slope of the velocity - time graph. On \([0,1]\), \(a>0\); on \([1,4]\), \(a = 0\); on \([4,5]\), \(a<0\); on \([5,7]\), \(a>0\); on \([7,10]\), \(a = 0\).

Answer:

a. \(t = 5\)
b. \(1\leq t\leq4,7\leq t\leq10\)
c. (Graphing requires a visual medium. Sketch a graph where speed is non - negative, increasing from \(0\) to a positive value on \([0,1]\), constant on \([1,4]\), decreasing to \(0\) on \([4,5]\), increasing from \(0\) to a negative value on \([5,7]\), and constant on \([7,10]\))
d. (Graph acceleration as a piece - wise function: positive on \([0,1]\), \(0\) on \([1,4]\) and \([7,10]\), negative on \([4,5]\), positive on \([5,7]\))