Question

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 1

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

Step-by-step 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

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

`39=10.4t+(a)/(2) *3.3^2`

`5.445a=39-34.32`

`a=39-34.32/5.445`

`a=4.68/5.445`

`a=0.86m/s^2`