To solve this problem we will apply the linear motion kinematic equations. For this purpose we know that the squared change of the initial velocity and the final velocity is equal to twice the acceleration by the distance traveled. Mathematically this can be expressed as,
[tex]v_f^2 - v_i^2 =2ax[/tex]
Where,
[tex]v_f[/tex] = Final velocity
[tex]v_i[/tex] = Initial velocity
[tex]a[/tex] = Acceleration
[tex]x[/tex] = Displacement
Replacing,
[tex](80)^2 - (40)^2= 2a(200)[/tex]
[tex]a = \frac{(80)^2 - (40)^2}{2(200)}[/tex]
[tex]a = 12m/s^2[/tex]
Therefore the acceleration during this time is [tex]12m/s^2[/tex]
