Determine the acceleration of the object during this motion

Question

An object moving on the x axis with a constant

acceleration increases its x coordinate

by 196 m in a time of 3.85 s and has a velocity

of 20 m/s at the end of this time. Answer in units of m/s^2

Answer 1

Assume it starts from x=0 at t=0 with initial speed vi and reaches x=196 m at t=3.85 s with final speed vf=20 m/s.

 

a = (vf - vi) / t = (20 - vi) / (3.85) = 5.195 - 0.260 * vi

 

Equation 1:  a = 5.195 - 0.260*vi

 

x(t) = vi*t + (1/2)*a*t^2 

 

x(3.85) = 196 = 3.85*vi + [(1/2)(3.85)^2]*a = 3.85*vi + 7.411*a

 

Equation 2:  196 = 3.85*vi + 7.411*a

 

Equations 1 and 2 are two equations in two unknowns.  Get vi in terms of a from equation 2 and plug into equation 1 to get a:

 

From equation 2:  vi = (196 - 7.411*a) / 3.85 = 50.909 - 1.925*a

 

Then from equation 1:  a = 5.195 - 0.260*(50.909 - 1.925*a)

 

a = -16.1 m/s^2