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

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

Step2: Identify constant - speed intervals

Speed is the absolute - value of velocity. The body moves at a constant speed when the magnitude of velocity is constant. From the graph, the body moves at a constant speed in the interval \(5\leq t\leq8\) (since \(v = 5\) m/s in this interval).

Step3: Graph speed

Speed \(s(t)=\vert v(t)\vert\). For \(0\leq t\leq2\), \(v(t)\) is increasing from \(0\) to \(5\), so \(s(t)\) is also increasing from \(0\) to \(5\). For \(2\leq t\leq4\), \(v(t)\) is decreasing from \(5\) to \(- 5\), so \(s(t)\) is decreasing from \(5\) to \(0\) and then increasing from \(0\) to \(5\). For \(4\leq t\leq5\), \(v(t)\) is increasing from \(-5\) to \(5\), so \(s(t)\) is decreasing from \(5\) to \(0\) and then increasing from \(0\) to \(5\). For \(5\leq t\leq8\), \(v(t)=5\), so \(s(t) = 5\). For \(8\leq t\leq10\), \(v(t)\) is decreasing from \(5\) to \(0\), so \(s(t)\) is decreasing from \(5\) to \(0\).

Step4: Graph acceleration

Acceleration \(a(t)=\frac{dv}{dt}\). For \(0\leq t\leq2\), \(a(t)\) is positive and constant (since \(v(t)\) is a straight - line with positive slope). For \(2\leq t\leq4\), \(a(t)\) is negative and constant (since \(v(t)\) is a straight - line with negative slope). For \(4\leq t\leq5\), \(a(t)\) is positive and constant. For \(5\leq t\leq8\), \(a(t) = 0\) (since \(v(t)\) is constant). For \(8\leq t\leq10\), \(a(t)\) is negative and constant.

Answer:

a. \(t = 2,4\)
b. \(5\leq t\leq8\)
c. Sketch a graph of speed as described in Step 3.
d. Sketch a graph of acceleration as described in Step 4.