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 body’s speed for 0 ≤ t ≤ 10. d. graph the acceleration, where defined.

Explanation:

Step1: Identify direction - change points

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

Step2: Identify constant - speed intervals

Constant speed means \(|v|\) is constant. From the graph, \(|v|\) is constant on the intervals \([0,1]\), \([4,6]\) and \([6,8]\).

Step3: Graph the speed

Speed \(s(t)=|v(t)|\). For \(0\leq t\leq1\), \(s(t)=v(t)\) (since \(v\geq0\)), for \(1 < t<2\), \(s(t)=v(t)\), for \(2 < t<4\), \(s(t)= - v(t)\) (since \(v<0\)), for \(4\leq t\leq6\), \(s(t)=v(t)\), for \(6\leq t\leq8\), \(s(t)=v(t)\) and for \(8 < t\leq10\), \(s(t)= - v(t)\).

Step4: Graph the acceleration

Acceleration \(a(t)=v^\prime(t)\). On \([0,1]\), \(a(t)\) is positive (linear - increasing velocity), on \((1,2)\), \(a(t)\) is negative, at \(t = 2\), \(a(t)\) has a non - differentiable point (sharp turn), on \((2,3)\), \(a(t)\) is negative, on \((3,4)\), \(a(t)\) is positive, on \([4,6]\), \(a(t) = 0\) (constant velocity), on \((6,8)\), \(a(t)=0\), on \((8,10)\), \(a(t)\) is negative.

Answer:

a. \(t = 2,4\)
b. \(t\in[0,1]\cup[4,6]\cup[6,8]\)
c. Graph speed by taking absolute - value of \(v(t)\) values for each \(t\) in \([0,10]\).
d. Graph acceleration by finding the slope of \(v(t)\) at each differentiable point in \([0,10]\) and note non - differentiable points.