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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?

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

Answer :

OKPALAWALTER8GO

Answer:

c)X= Vo(t)+1/2at^2

Explanation:

From the question we are told that

initial speed u=10.4

displacement d= 39

time t=3.3s

accelaration a=?

Generally the best equation in solving for Acceleration is given to be

X= Vo(t)+1/2at^2

Mathematically

     [tex]X= Vo(t)+1/2at^2[/tex]

     [tex]39=10.4t+\frac{a}{2} *3.3^2[/tex]

     [tex]5.445a=39-34.32[/tex]

     [tex]a=39-34.32/5.445[/tex]

     [tex]a=4.68/5.445[/tex]

     [tex]a=0.86m/s^2[/tex]

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