An object moves with constant acceleration 3.10 m/s2 and over a time interval reaches a final velocity of 12.4 m/s. If its initial velocity is - 6.20 m/s, what is its displacement during this interval?

Given:

a = 3.10 m/s^2

vf = 12.4 m/s

vi = - 6.2 m/s

t = (vf - vi) / a

= (12.4 + 6.2) / 3.1

= 6 s

displacement = (vf - vi) * t

= (12.4 + 6.2) * 6

= 111.6 m.

25.05 m

Explanation:

Using newton's equation of motion.

v² = u²+2as ... Equation 1

Where v = final velocity of the object, u = initial velocity of the object, a = acceleration of the object, s = displacement of the object.

make s the subject of the equation

s = (v²-u²) / 2a ... Equation 2

Given: v = 12.4 m/s, u = - 6.20 m/s, a = 3.10 m/s²

Substitute into equation 2

s = (12.4² - (-6.2) ²) / (2*3.1)

s = (153.76-38.44) / 6.2

s = 155.32/6.2

s = 25.05 m.

Hence the displacement = 25.05 m