La trayectoria de vuelo del helicóptero cuando despega desde A esta definida por las ecuaciones x=2t^2 m y y=0.04t^3 m, donde t es el tiempo en segundos. Determine la distancia a la que está el helicóptero desde el punto A y las magnitudes de su velocidad y su aceleración cuando t=10 s

The flight path of the helicopter when it takes off from A is defined by the equations x = 2t ^ 2 m and y = 0.04t ^ 3 m, where t is the time in seconds. Determine the distance the helicopter is from point A and the magnitudes of its speed and acceleration when t = 10 s

Explanation:

Given that,

The path curve are

x = 2t²

y = 0.04t³

Distance from A. When t = 10s

The position on the x axis at t=10s is

x = 2t²

x = 2×10² = 200m

The position on y axis at t= 10s is

y = 0.04t³

y = 0.04 ×10³ = 40m

So, the plane it at (200,40) and it started for (0,0)

From geometry we can find distance between the two point using

D = √(x2-x1)² + (y2-y1)²

D = √(200-0)²+(40-0)²

D = √200²+40²

D = 203.96m

So, the distance is approximately 204m

b. The magnitude of the speed.

Speed is give as the differentiation of distance

So, for x curve, at t =10

Vx = dx/dt = 4t

Vx = 4×10 = 40m/s

Also, for y axis at t=10

Vy = dy/dt = 0.12t²

Vy = 0.12×10² = 12m/s

Then, the velocity at 10s is

V = √(Vx²+Vy²)

V = √40²+12²

V = 41.76m/s

The velocity helicopter at t=10s is 41.76m/s

c. The acceleration at t=10s?

We need to find acceleration of the curve and it is given as

a = d²x/dt²

So, for x axis at t=10

ax= d²x/dt² = 4

ax = 4m/s²

Also, for y axis at t=10

ay= d²y/dt² = 0.24t

ay = 0.24 ×10 = 2.4m/s²

Then, the magnitude of acceleration at t=10s is

a = √(ay²+ax²)

a = √2.4²+ 4²

ay = 4.66m/s²

The acceleration of the helicopter at t=10s is 4.66m/s²

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