The jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point a

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

         = 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

          = 0.8

at x= 5 Km

dy/dx = 0.8(5)

         = 4

now insert the values,

→ Now the normal component of acceleration is given by

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = 0.921 m/s²

answer is c

explanation:

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