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,
[tex]a_{t} = v = 0.8 m/s^{2}[/tex]
→ 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
[tex]p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }[/tex]
now insert the values,
[tex]p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km[/tex]
→ Now the normal component of acceleration is given by
[tex]a_{n} = \frac{v^{2} }{p}[/tex]
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
[tex]a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}[/tex]
[tex]a = [(0.8)^{2} + (0.457)^{2}]^{0.5}[/tex]
a = 0.921 m/s²
