Question

problem 7. an arrow is launched with an initial velocity of 3.2 m/s. it travels a displacement of 60 m in 10 s. calculate the final velocity of the arrow.
problem 8. a sponge is knocked off a table with an initial velocity of 1.3 m/s. it accelerates at a rate of 10 m/s² until it reaches a final velocity of 55 m/s. calculate the amount of time that the sponge was falling.

Explanation:

Problem 7

Step1: Identify the kinematic - equation

We use the equation $d = v_0t+\frac{1}{2}at^2$ and $v = v_0 + at$. First, from $d = v_0t+\frac{1}{2}at^2$, we can find acceleration $a$. Given $v_0 = 3.2\ m/s$, $d = 60\ m$, $t = 10\ s$.
$60=3.2\times10+\frac{1}{2}a\times(10)^2$

Step2: Solve for acceleration $a$

$60 = 32+50a$
$50a=60 - 32$
$50a = 28$
$a=\frac{28}{50}=0.56\ m/s^2$

Step3: Calculate the final velocity $v$

Use $v = v_0+at$. Substitute $v_0 = 3.2\ m/s$, $a = 0.56\ m/s^2$, $t = 10\ s$
$v=3.2+0.56\times10$
$v=3.2 + 5.6$
$v = 8.8\ m/s$

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration and $t$ is the time. We need to solve for $t$.
Given $v_0 = 1.3\ m/s$, $a = 10\ m/s^2$, $v = 55\ m/s$

Step2: Rearrange the equation to solve for $t$

$v=v_0 + at$ can be rewritten as $t=\frac{v - v_0}{a}$

Step3: Substitute the values

$t=\frac{55 - 1.3}{10}$
$t=\frac{53.7}{10}=5.37\ s$

Answer:

$8.8\ m/s$

Problem 8