Question

A firefighter slides down a pole in a fire station with an acceleration of 8.1 m/s2 .
If the firefighter starts at rest and slides for a distance of 3.7 m , what is his final speed?

Answer 1

Answer:

The firefighter final speed is 7.74 meters per second (m/s)

Explanation:

Uniformly Accelerated Motion

We are asked for the firefighter final speed

Since we know his initial velocity, the acceleration and the distance traveled,

The most suitable motion equation to apply is:

$\large\boxed {\bold {(V_{f})^{2} = (V_{0})^{2} + 2 \ . \ a \ .\ d }}$

$\bold { V_{f} } \ \ \ \ \ \ \textsf{ Final Velocity }\ \ \ \bold{ \frac{m}{s} }$

$\bold { V_{0}} \ \ \ \ \ \ \textsf{ Initial Velocity }\ \ \ \bold{0 \ \frac{m}{s} }$

$\bold { a }\ \ \ \ \ \ \ \ \textsf{ Acceleration }\ \ \ \bold{8.1 \ \frac{m}{s^{2} } }$

$\bold { d} \ \ \ \ \ \ \ \ \textsf{ Distance }\ \ \ \bold{3.7 \ m }$

As the firefighter starts at rest his initial velocity wlll be zero = 0 $\bold {V_{0} = 0 }$

Thus we have

$\boxed {\bold {(V_{f})^{2} = (V_{0})^{2} + 2 \ . \ a \ .\ d }}$

$\bold {V_{0} = 0 }$

$\boxed {\bold {(V_{f})^{2} = 0 + 2 \ . \ a \ .\ d }}$

$\boxed {\bold {(V_{f})^{2} = 2 \ . \ a \ .\ d }}$

$\boxed {\bold { \sqrt{ (V_{f})^{2} } =\sqrt{ 2 \ . \ a \ .\ d } }}$

$\large\boxed {\bold {V_{f} =\sqrt{ 2 \ . \ a \ .\ d } }}$

$\large\textsf{ Substituting the given values and solving }$

$\boxed {\bold {V_{f} =\sqrt{ 2 \ . \ 8.1 \ \frac{m}{s^{2} } \ .\ 3.7\ m } }}$

$\boxed {\bold {V_{f} =\sqrt{ 59.94\ \frac{m^{2} }{s^{2} } } }}$

$\boxed {\bold {V_{f} = 7.74209 \ \frac{m}{s} }}$

$\large\boxed {\bold {V_{f} = 7.74 \ \frac{m}{s} }}$

Hence the firefighter final speed is 7.74 meters per second (m/s)