An airplane traveling north at 400 m/s is accelerated due east at a rate of

50 m/s2 for 6 s. If the effects of air resistance and gravity are ignored, what is
the final speed of the plane?

50 m/s^2 * 6s = 300 m/s

v = 400j + 300i
speed = 500

To find the final speed of the plane, we can use the concept of vector addition. Since the airplane is traveling north at a constant speed, we can represent its initial velocity as a vector pointing straight up:

Initial velocity (v1) = 400 m/s, pointing north

The acceleration due east can be represented as a vector pointing to the right:

Acceleration (a) = 50 m/s^2, pointing east

To find the final velocity, we need to add the change in velocity to the initial velocity:

Final velocity (v2) = v1 + Δv

Now, we can calculate the change in velocity (Δv) using the formula:

Δv = a × t

where t is the time taken for acceleration, which is 6 seconds.

Δv = 50 m/s^2 × 6 s
Δv = 300 m/s, pointing east

Adding the change in velocity to the initial velocity gives us the final velocity:

Final velocity (v2) = 400 m/s (north) + 300 m/s (east)

To add these vectors, we can use the Pythagorean theorem because they are perpendicular to each other:

v2 = √[(400 m/s)^2 + (300 m/s)^2]

Calculating this:

v2 = √[(160,000 m^2/s^2) + (90,000 m^2/s^2)]
v2 = √[250,000 m^2/s^2]
v2 = 500 m/s

Therefore, the final speed of the plane is 500 m/s.

To solve this problem, we can use the principles of vector addition. We will break down the airplane's velocity into two components: one in the north direction, and one in the east direction.

Given:
Initial velocity in the north direction (Vnorth) = 400 m/s
Acceleration in the east direction (Aeast) = 50 m/s^2
Time (t) = 6 seconds

First, we need to calculate the change in velocity in the east direction (ΔVeast) using the formula:

ΔVeast = Aeast * t

ΔVeast = 50 m/s^2 * 6 s
ΔVeast = 300 m/s

Next, we can use the Pythagorean theorem to find the magnitude of the final velocity (Vfinal), combining the components in the north and east direction:

Vfinal = √(Vnorth^2 + ΔVeast^2)

Vfinal = √(400 m/s)^2 + (300 m/s)^2
Vfinal = √(160000 m^2/s^2 + 90000 m^2/s^2)
Vfinal = √(250000 m^2/s^2)
Vfinal = 500 m/s

Therefore, the final speed of the plane is 500 m/s.