Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

Explanation:

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.