Question

Nathan accelerates his skateboard uniformly along a straight path from rest to 12.5 m/s in 2.5 s.
a. What is Nathanâs acceleration?
b. What is Nathanâs displacement during this time interval?
c. What is Nathanâs average velocity during this time interval?

Answer 1

Answer:

a) Nathan's acceleration is 5 m/s²

b) Nathan's displacement during this time interval is 15.625 m

c) Nathan's average velocity during this time interval is 6.25 m/s

Explanation:

a) We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 12.5 m/s    

     Time, t = 2.5 s

     Substituting

                      v = u + at  

                      1.25 = 0 + a x 2.5

                      a = 5 m/s²

     Nathan's acceleration is 5 m/s²

b) We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 5 m/s²  

        Time, t = 2.5 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 2.5 + 0.5 x 5 x 2.5²

                      s = 15.625 m

      Nathan's displacement during this time interval is 15.625 m

c) Displacement = 15.625 m

   Time = 2.5 s

  We have

           Displacement = Time x Average velocity

           15.625 = 2.5 x  Average velocity

           Average velocity = 6.25 m/s

     Nathan's average velocity during this time interval is 6.25 m/s

Answer 2

A. Nathan's acceleration is 5 m/s²

B. Nathan's displacement during the time interval is 15.625 m

C. Nathan's average velocity during the time interval is 6.25 m/s

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

Initial velocity (u) = 0 m/s

Final velocity (v) = 12.5 m/s

Time (t) = 2.5 s

A. Determination of the Nathan's acceleration.

Acceleration is defined as the rate of change of velocity with time. $Acceleration = \frac{final velocity - initial velocity}{time} \\\\a = \frac{v - u}{t}$

With the above formula, we can obtain Nathan's acceleration as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12.5 m/s

Time (t) = 2.5 s

Acceleration (a) =?

$a = \frac{v - u}{t} \\\\a = \frac{12.5 - 0}{2.5} \\\\a = \frac{12.5 }{2.5}$

a = 5 m/s²

Thus, Nathan's acceleration is 5 m/s²

B. Determination of Nathan's displacement.

Displacement is defined as the distance travelled in a specific direction.

Nathan's displacement can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12.5 m/s

Time (t) = 2.5 s

Displacement (s) =?

$s = \frac{(v + u)t}{2}\\\\s = \frac{(12.5 + 0)2.5}{2}\\\\s = \frac{(12.5)2.5}{2}\\\\s =\frac{31.25}{2}$

s = 15.625 m

Therefore, Nathan's displacement is 15.625 m

C. Determination of Nathan's average velocity.

Average velocity is defined as the total displacement travelled divided by the total time.

K$Average velocity = \frac{Total displacement }{time}$

With the above formula, we can obtain the average velocity of Nathan as follow:

Total displacement = 15.625 m

Total time = 2.5 s

Average velocity =.?

$Average velocity = \frac{Total displacement}{total time }\\\\Average velocity = \frac{15.625}{2.5}$

Average velocity = 6.25 m/s

Therefore, Nathan's average velocity is 6.25 m/s

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