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12345 [234]
3 years ago
9

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 t

he rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point a
Physics
1 answer:
givi [52]3 years ago
4 0

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,

a_{t} = v = 0.8 m/s^{2}

→ 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

p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} }   }{\frac{d^{2y} }{dx^{2} } }

now insert the values,

p = \frac{[1+(4)^{2}]^{\frac{3}{2} }  }{0.8}  = 87.62 km

→ Now the normal component of acceleration is given by

a_{n} = \frac{v^{2} }{p}

    = (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}

a = [(0.8)^{2} + (0.457)^{2}]^{0.5}

a = 0.921 m/s²

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