6. A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of 42m/s What is the approximate acceleration of the train during this time?

0.17 m/s2

0.20 m/s2

0.16 m/s2

0.14 m/s2

0.19 m/s2

O. 16 m/s2

Answer;

0.16 m/s²

Explanation;

From one of the equation of linear motion;

vf2 - vi2 = 2ax

Where;

vf = final speed after traveling a distance x at constant acceleration; vf = 42 m/s

vi = initial speed = 0 (starts from rest)

a = constant acceleration while traveling a distance x

x = distance traveled at constant acceleration; x = 5.6 km = 5600 m

Therefore;

a = vf2 / (2x)

= 422 / (2*5600) m/s²

= 0.1575 m/s²

≅ 0.16 m/s²