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.
the body reverses direction at t = (use a comma to separate answers as needed )

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

Speed is the absolute - value of velocity. The body is moving at a constant speed when the magnitude of velocity is constant. From the graph, this occurs for $4\leq t\leq8$.

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 $4\leq t\leq8$, $v(t)=5$, 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 $a(t)=v^\prime(t)$. 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. For $4\leq t\leq8$, $a(t) = 0$ (since $v(t)$ is constant). For $8\leq t\leq10$, $a(t)$ is negative and constant.

Answer:

a. $t = 2,4$
b. $4\leq t\leq8$
c. Graph speed by taking the absolute - value of the velocity values at each $t$ in the interval $0\leq t\leq10$.
d. Graph acceleration by finding the slope of the velocity function at each $t$ where the slope is well - defined.