For a short period of time a jet aircraft follows the path y

For a short period of time a jet aircraft follows the path y= 0.00000125x ft . When the aircraft is located at (x,y)=(1000,1250) ft , it has a speed of 300 mph which is increasing at a rate of 12ft/s^2 . Determine the magnitude of the aircraft’s acceleration at this instant.

Solution

Let it velocity at this point be:

v=vxi+vyj

Magnitude of v is 300 mph

sqrt{vx^2+vy^2}=300

vx=dx/dt,dy=dy/dt

vy/vx==dy/dx=0.00000125

vy=0.00000125vx

300=vx\\sqrt{1+0.00000125^2}

Solving gives:

vx~300 ,vy~=0

acceleration ,a=dv/dt=dvx/dti+dvy/dtj=(i+0.00000125j)dvx/dt~dvx/dt i

d^2vx/dt^2~12 ft/s^2

Hence, a=12 ft/s^2