An object moves with constant acceleration 4.20 m/s2 and over a time interval reaches a final velocity of 10.4 m/s.
(a) If its original velocity is 5.20 m/s, what is its displacement during the time interval?
V^2 = Vo^2 + 2a*d.
V = 10.4 m/s, Vo = 5.20 m/s, a = 4.20 m/s, d = ?.
To find the displacement of the object during the time interval, we need to use the equations of motion.
First, let's list down the given values:
Initial velocity (u) = 5.2 m/s
Acceleration (a) = 4.2 m/s^2
Final velocity (v) = 10.4 m/s
We can use the equation of motion to relate displacement, initial velocity, final velocity, and acceleration:
v^2 = u^2 + 2aS
where v is the final velocity, u is the initial velocity, a is the acceleration, and S is the displacement.
Rearrange the equation to solve for S:
S = (v^2 - u^2) / (2a)
Now, substitute the given values into the equation:
S = (10.4^2 - 5.2^2) / (2 * 4.2)
Calculate the expression:
S = (108.16 - 27.04) / 8.4
S = 81.12 / 8.4
S = 9.64 meters
Therefore, the object's displacement during the time interval is 9.64 meters.