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.\na. when does the body reverse direction?\nb. when is it moving at a constant speed?\nc. graph the bodys speed for 0 ≤ t ≤ 10.\nd. graph the acceleration, where defined.\nthe body reverses direction at t = 2,4 (use a comma to separate answers as needed.)\nthe body moves at a constant speed at 5 ≤ t ≤ 8\nchoose the correct graph of the bodys speed for 0 ≤ t ≤ 10\n

the accompanying figure shows the velocity v = \\(\\frac{ds}{dt}=f(t)\\) (m/sec) of a body moving along a coordinate line.\na. when does the body reverse direction?\nb. when is it moving at a constant speed?\nc. graph the bodys speed for 0 ≤ t ≤ 10.\nd. graph the acceleration, where defined.\nthe body reverses direction at t = 2,4 (use a comma to separate answers as needed.)\nthe body moves at a constant speed at 5 ≤ t ≤ 8\nchoose the correct graph of the bodys speed for 0 ≤ t ≤ 10\n

Answer

Explanation:

Step1: Identify direction - change points

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 has a constant speed when the magnitude of velocity is constant. From the graph, (|v|) is constant for (5\leq t\leq8).

Step3: Graph the speed

Speed (s(t)=|v(t)|). For (0\leq t\leq2), (v(t)\geq0), so (s(t)=v(t)); for (2 < t<4), (v(t)<0), so (s(t)= - v(t)); for (4 < t\leq5), (v(t)\geq0), so (s(t)=v(t)); for (5\leq t\leq8), (v(t)) is constant, so (s(t)) is constant; for (8 < t\leq10), (v(t)) is decreasing, so (s(t)=|v(t)|). The correct graph of speed is the one where the values are non - negative (since speed is non - negative) and match the magnitude of the velocity graph. The correct graph is C as speed is always non - negative and has constant values when velocity is constant in magnitude.

Step4: Graph the acceleration

Acceleration (a(t)=\frac{dv}{dt}). The acceleration is the slope of the velocity - time graph. For (0 < t<2), (a(t)>0); at (t = 2), (a(t)) has a non - continuous change (a sharp turn in the velocity graph); for (2 < t<4), (a(t)<0); at (t = 4), (a(t)) has a non - continuous change; for (4 < t<5), (a(t)>0); for (5 < t<8), (a(t)=0) (since (v) is constant); for (8 < t<10), (a(t)<0).

Answer:

a. (t = 2,4) b. (5\leq t\leq8) c. C d. (Graph of acceleration: positive slope for (0 < t<2), negative slope for (2 < t<4), positive slope for (4 < t<5), zero slope for (5 < t<8), negative slope for (8 < t<10) with non - continuous changes at (t = 2) and (t = 4))