Answer:
[tex] X= v_i t + \frac{1}{2}a t^2 [/tex]
And from this equation we can solve for a like this:
[tex] 205m = 3.5m/s *(28.7s) +\frac{1}{2}a (28.7s)^2[/tex]
And solving for a we got:
[tex] 104.55m = \frac{1}{2}a (28.7s)^2[/tex]
[tex] a = \frac{2*104.55m}{(28.7s)^2)}= 0.254 m/s^2[/tex]
Step-by-step explanation:
For this case we have the velocity , distance and time given:
[tex] v = 3.5 m/s, d=205m, t =28.7s[/tex]
And we know from kinematics that he velocity can be expressed like this:
[tex] v_f = v_i +a t[/tex]
We also know that the distance is given by:
[tex] X= v_i t + \frac{1}{2}a t^2 [/tex]
And from this equation we can solve for a like this:
[tex] 205m = 3.5m/s *(28.7s) +\frac{1}{2}a (28.7s)^2[/tex]
And solving for a we got:
[tex] 104.55m = \frac{1}{2}a (28.7s)^2[/tex]
[tex] a = \frac{2*104.55m}{(28.7s)^2)}= 0.254 m/s^2[/tex]
