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A skateboarder traveling at 10.4 m/s starts getting chased by a grumpy dog. The skateboarder speeds up with a constant acceleration for 39m over a 3.3s. We want to find the acceleration of the skateboarder over the 3.3s time interval.

Which kinematic formula would be most useful to solve for the target unknown?

Picture shows answer choices better.
A. V= Vo+a(t)
B. X= (V+Vo/2)t
C. X= Vo(t)+1/2at^2
D. V^2 = Vo^2 +2ax
E. X= vt-1/2at^2

A skateboarder traveling at 104 ms starts getting chased by a grumpy dog The skateboarder speeds up with a constant acceleration for 39m over a 33s We want to f class=

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The answer is C.......

Answer:

[tex]x=v_ot+\dfrac{1}{2}at^2[/tex]                                                  

Explanation:

It is given that,

Initial speed of the skateboarder, [tex]v_o=10.4\ m/s[/tex]

It speeds up for 39 meters over a 3.3 seconds. Let a is the acceleration of the skateboarder over the 3.3 s time interval. To find the acceleration of the skateboarder we can use the second equation of motion. The kinematic formula that can be used is as follows :

[tex]x=v_ot+\dfrac{1}{2}at^2[/tex]

Here, x = 39 meters

[tex]v_o=10.4\ m/s[/tex]

t = 3.3 seconds

By rearranging the above equation, we can find the value of a. So, the correct option is (c).