An object moves with constant acceleration 4.95 m/s2 and over a time interval reaches a final velocity of 13.2 m/s.

(a) If its initial velocity is 6.6 m/s, what is its displacement during the time interval?

v = at

13.2 = 4.95t
t = 8/3

s = vt + 1/2 at^2
s = 6.6*8/3 + 1/2 * 4.95 * (8/3)^2
s = 35.2

a) 13.2

13.2^2=6.6^2+2(4.95)(ΔX)
174.24= 43.56+9.9ΔX
130.68=9.9ΔX
ΔX=13.2

To find the displacement of the object during the time interval, we can use the formula:

displacement = (final velocity)^2 - (initial velocity)^2 / (2 * acceleration)

Given:
Initial velocity (u) = 6.6 m/s
Final velocity (v) = 13.2 m/s
Acceleration (a) = 4.95 m/s²

Substituting the values into the formula, we have:

displacement = (13.2^2 - 6.6^2) / (2 * 4.95)

Calculating the values inside the parentheses first:

displacement = (174.24 - 43.56) / (2 * 4.95)

displacement = 130.68 / 9.9

Finally, calculating the division:

displacement = 13.2 meters

Therefore, the displacement of the object during the time interval is 13.2 meters.