the accompanying figure shows the velocity v = $\frac{ds}{dt}$ = f(t) (m/sec) of a body moving along a…

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.

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.

Answer

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.