Question

An object with an initial speed of 4.0 meters per second accelerates uniformly at 2.0 meters per second squared in the direction of its motion for a distance fof 5 meters. What is the final speed of the object

Answer 1

Answer:

6 m/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 4 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

The final velocity of the object can obtained as shown below:

v² = u² + 2as

v² = 4² + (2 × 2 × 5)

v² = 16 + 20

v² = 36

Take the square root of both side.

v = √36

v = 6 m/s

Therefore, the final speed of the object is 6 m/s.

Answer 2

The final speed of the object as it accelerates uniformly in the direction of its motion is 6m/s.

Given the data in the question;

• Initial speed; $u = 4.0m/s$

• Acceleration; $a = 2.0m/s^2$

• Distance; $s = 5m$

Final speed; $v = \ ?$

To determine the final speed of the object, we use the third equation of motion:

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

Where v is the final speed or velocity, u is the initial velocity, a is the acceleration and s is the distance covered.

We substitute our given values into the equation

$v^2 = [(4.0m/s)^2] + [ 2\ *\ 2.0m/s^2\ *\ 5m ]\\\\v^2 = 16m^2/s^2 + 20m^2/s^2\\\\v^2 = 36m^2/s^2\\\\v = \sqrt{36m^2/s^2} \\\\v = 6m/s$

Therefore, the final speed of the object as it accelerates uniformly in the direction of its motion is 6m/s.

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