a bubble be jump straight up with a velocity of 4.0 m/s what is his displacement of after 1.80s
displacement=velocity*time
To find the displacement of the bubble after 1.80 seconds, we need to use the equation of motion for uniformly accelerated motion:
displacement = initial velocity × time + (1/2) × acceleration × time^2
In this case, since the bubble is jumping straight up, its acceleration is equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it is acting in the opposite direction of the initial velocity).
Given:
Initial velocity (u) = 4.0 m/s
Time (t) = 1.80 s
Acceleration (a) = -9.8 m/s^2
Plugging these values into the equation, we get:
displacement = (4.0 m/s) × (1.80 s) + (1/2) × (-9.8 m/s^2) × (1.80 s)^2
Now, let's calculate it step by step:
Step 1: Calculate (1/2) × (-9.8 m/s^2) × (1.80 s)^2
= (1/2) × (-9.8 m/s^2) × (3.24 s^2)
= -15.1164 m
Step 2: Calculate (4.0 m/s) × (1.80 s)
= 7.2 m
Step 3: Add the results from Step 1 and Step 2 to get the displacement:
displacement = 7.2 m + (-15.1164 m)
= -7.9164 m
Therefore, the displacement of the bubble after 1.80 seconds is approximately -7.92 m. The negative sign indicates that the bubble is moving in the opposite direction of its initial velocity.