Question

How many seconds are required for the ranger’s vehicle in the previous problem to stop? Only enter the number not the units

Answer 1

Answer:

distance = 17.1 m

time = 3 s

Explanation:

First, we will use the following equation:

$v^2_f=v^2_i+2a(\Delta x)$

Where vf is the final velocity, vi is the initial velocity, a is the acceleration and

Δx is the distance. Then, replacing each value, we get:

$\begin{gathered} 0^2=11.4^2+2(-3.8)(\Delta x) \\ 0=129.96-7.6\Delta x \end{gathered}$

The final velocity is 0 m/s because the vehicle will come to rest and the acceleration is negative because it is slowing down.

Then, solving for Δx, we get:

$\begin{gathered} 7.6\Delta x=129.96 \\ \frac{7.6\Delta x}{7.6}=\frac{129.96}{7.6} \\ \Delta x=17.1\text{ m} \end{gathered}$

Now, with the distance and velocities, we can find the time t using the following equation:

$\Delta x=\frac{1}{2}(v_i+v_f)t$

So, replacing the values and solving for t, we get:

$\begin{gathered} 17.1=\frac{1}{2}(11.4+0)t \\ 17.1=\frac{1}{2}(11.4)t \\ 17.1=5.7t \\ \frac{17.1}{5.7}=\frac{5.7t}{5.7} \\ 3\text{ s = t} \end{gathered}$

Therefore, the answers are:

distance = 17.1 m

time = 3 s