Question

A firefighter slides down a pole in a fire station with an acceleration of 7.8 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 final speed of a firefighter is 7.59 m/s.        

Explanation:

Given that,

Acceleration of a firefighter, $a=7.8\ m/s^2$

Initial speed of firefighter, u = 0 (at rest)

Distance covered, d = 3.7 m

Let v is his final speed. using third equation of motion to find it as :

$v^2-u^2=2ad\\\\v^2=2ad\\\\v=\sqrt{2\times 7.8\times 3.7} \\\\v=7.59\ m/s$

So, the final speed of a firefighter is 7.59 m/s.

Answer 2

The final speed of the firefighter is 7.59 m/s.

Given data:

The magnitude of acceleration of firefighter is, $a = 7.8 \;\rm m/s^{2}$.

The initial speed of firefighter is, u = 0 (Since, the firefighter starts from rest).

The sliding distance is, s = 3.7 m.

The given problem is based on the kinematic equations of motion. The third kinematic equation of motion can be used to obtain the final speed of firefighter. The third kinematic equation of motion is given as,

$v^{2} = u^{2} + 2as$

Here, v is the final speed of firefighter.

Solve by substituting the values as,

$v^{2} = 0^{2} +( 2 \times 7.8 \times 3.7)\\\\v=\sqrt{2 \times 7.8 \times 3.7}\\\\v = 7.59 \;\rm m/s$

Thus, we can conclude that the final speed of the firefighter is 7.59 m/s.

Learn more about the kinematic equations of motion here:

https://brainly.com/question/14355103