To determine the distance traveled by a train in the sixth second, we need to use the equations of motion. In this case, we have an object starting from rest (v0=0v_0 = 0v0=0) with a uniform acceleration (a=4.0 m/s2a = 4.0 , ext{m/s}^2a=4.0m/s2).
The equation that relates distance (ddd), initial velocity (v0v_0v0), acceleration (aaa), and time (ttt) is:
d=v0⋅t+12⋅a⋅t2d = v_0 cdot t + frac{1}{2} cdot a cdot t^2d=v0⋅t+21⋅a⋅t2
Plugging in the values:
v0=0 m/sv_0 = 0 , ext{m/s}v0=0m/s a=4.0 m/s2a = 4.0 , ext{m/s}^2a=4.0m/s2 t<mo