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.
the body reverses direction at t = 2.4 (use a comma to separate answers as needed.)
the body moves at a constant speed at 5≤t≤8
choose the correct graph of the bodys speed for 0≤t≤10.

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, $|v| = 5$ for $5\leq t\leq8$.

Step3: Graph the speed

Speed is $|v|$. For $0\leq t\leq2$, $v$ is non - negative, so speed $s=v$. For $2 < t<4$, $v$ is negative, so $s=-v$. For $4 < t\leq5$, $v$ is non - negative, so $s = v$. For $5\leq t\leq8$, $|v| = 5$. For $8 < t\leq10$, $v$ is non - positive, so $s=-v$. The correct graph of speed is the one where the values match these conditions.

Step4: Graph the acceleration

Acceleration $a=\frac{dv}{dt}$. It is the slope of the velocity - time graph. For $0 < t<1$, $a>0$ (positive slope), for $1 < t<2$, $a < 0$ (negative slope), for $2 < t<3$, $a<0$, for $3 < t<4$, $a>0$, for $4 < t<5$, $a>0$, for $5 < t<8$, $a = 0$ (constant velocity), for $8 < t<10$, $a<0$.

Answer:

a. $t = 2,4$
b. $5\leq t\leq8$
c. The correct graph is the one that shows the absolute - value of the velocity function over the interval $0\leq t\leq10$. Analyzing the behavior of $v(t)$: when $v\geq0$, speed $s = v$ and when $v<0$, speed $s=-v$. The correct graph is C.
d. Graph the slope of the velocity - time graph over the intervals where the slope is well - defined.