The jet plane travels along the vertical parabolic path When

The jet plane travels along the vertical parabolic path. When it is at point A it has a speed of 200 m/s, which is increasing at the rate of 0.8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

Solution

GIVEN DATA:

SPEED OF HER PLANE V= 200 m/sec=200/1000=0.2 km/sec. HORIZONTAL DISTANCE x=5 kms. VERTICAL DISTANCE y=10 kms. TANGENTIAL ACCELERATION a_t= 0.8 m/sec².

PLANE A EQUATION y=0.4x²

TO FIND : MAGNITUDE OF ACCELERATION OF THE PLANE AT POINT A. SOLUTION: y=0.4x² differentiating y with respect to x....therefore dy/dx= 2×0.4x¹= 0.8x put x=5 kms dy/dx= 0.8×5=4....further integrating dy/dx, d²y/dx²=0.8×1=0.8 Displacement= (1+(dy/dx)²)^3/2/d²y/dx² = (1+4²)^3/2/0.8=87.62 kms. NORMAL ACCELERATION a_n= v²/Dispacement= 0.200²/87.62=0.457× 10^-3 km/sec²=0.457 m/sec².TANGENTIAL ACCELERATION a_t=0.8 m/sec²

MAGNITUDE OF ACCELERATION OF PLANE AT POINT A. a=(a_t² + a_n²)^1/2 = (0.8² + 0.457²)^1/2 = 0.921 m/sec².