Answer:
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
