A boat, having an initial velocity of 6.6m/s east undergoes an average acceleration of 2.2 m/s squared south for 2.5 s. Determine final velocity after the time interval (magnitute and direction)
Well, that boat is in for a wild ride! Let's see if I can calculate its final velocity using my comical calculations.
To determine the final velocity, we need to consider the boat's initial velocity, the average acceleration, and the time interval. So, here we go:
The initial velocity of the boat is 6.6 m/s east. Let's call it "Boaty McBoatface" just for fun.
The average acceleration is 2.2 m/s² south. That's like the boat suddenly wanting to take a nosedive into the sea!
The time interval is 2.5 seconds. That's shorter than a clown's attention span!
Now, to calculate the final velocity, we need to combine the initial velocity with the change in velocity caused by the acceleration.
Since the initial velocity is in the east direction and the acceleration is towards the south, they are perpendicular to each other. So, we can treat this problem as a right-angled triangle.
Using the Pythagorean theorem for the magnitude, we have:
Final velocity magnitude = √(initial velocity magnitude² + change in velocity magnitude²)
Final velocity magnitude = √(6.6 m/s²² + (2.2 m/s² * 2.5 s)²)
Final velocity magnitude = √(6.6 m/s)² + (13.75 m/s)²)
Final velocity magnitude ≈ √(43.56 m²/s² + 189.06 m²/s²)
Final velocity magnitude ≈ √(232.62 m²/s²)
Final velocity magnitude ≈ 15.25 m/s
Now, let's determine the direction. Since the initial velocity is east and the acceleration is south, the final velocity will be somewhere in between.
So, the final velocity for Boaty McBoatface after 2.5 seconds will be approximately 15.25 m/s in a south-east direction.
I hope you found this comical calculation helpful!
To determine the final velocity of the boat after the given time interval, we can use the equation:
v = u + a*t
Where:
v = final velocity
u = initial velocity
a = average acceleration
t = time interval
Given:
u = 6.6 m/s (east)
a = 2.2 m/s^2 (south)
t = 2.5 s
To solve for the final velocity, substitute the given values into the equation:
v = 6.6 m/s + (2.2 m/s^2) * (2.5 s)
v = 6.6 m/s + 5.5 m/s
v = 12.1 m/s
Therefore, the final velocity of the boat after the time interval is 12.1 m/s. Since the initial velocity was in the east direction and the acceleration was in the south direction, the final velocity will have a magnitude of 12.1 m/s, but the direction will be the resultant of the east and south components.
To determine the final velocity of the boat, we can use the equation of motion:
Vf = Vi + at
where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time interval
Given:
Vi = 6.6 m/s east
a = 2.2 m/s^2 south
t = 2.5 s
First, let's determine the change in velocity (ΔV) in the south direction using the formula:
ΔV = at
ΔV = 2.2 m/s^2 * 2.5 s = 5.5 m/s south
Since the boat moves east initially, in order to combine the east and south components of the velocity, we can use the Pythagorean theorem:
Vf^2 = (Ve)^2 + (Vs)^2
where:
Vf = final velocity
Ve = east component of velocity
Vs = south component of velocity
Given:
Ve (initial) = 6.6 m/s (east)
Rearranging the equation:
(Vf)^2 = (6.6 m/s)^2 + (5.5 m/s)^2
(Vf)^2 = 43.56 m^2/s^2 + 30.25 m^2/s^2
(Vf)^2 = 73.81 m^2/s^2
Taking the square root of both sides:
Vf = √73.81 m/s ≈ 8.60 m/s
The final velocity of the boat after the time interval is approximately 8.60 m/s. To determine the direction, we need to find the angle between the east direction and the resultant velocity. We can use the tangent function:
θ = tan^(-1) (Vs / Ve)
θ = tan^(-1) (5.5 m/s / 6.6 m/s)
θ ≈ 40.74°
Therefore, the final velocity of the boat after the time interval is approximately 8.60 m/s in a direction of 40.74° south of east.
Vf=Vi+at
= 6.6E+2.2*2.5 S
that is it, unless you want to convert to polar.