Ajet plane flying 600 m/s experiences an acceleration of 4g when pulling out of the dive. what is the radius of curvature of the loop in which the plane is flying?

R = 9200 m

Explanation:

given,

speed of the plane = 600 m/s

acceleration = 4 g

Radius of curvature of loop = ?

we know,

angular acceleration

.........(1)

we know,

v = r ω

putting value of r in equation (1)

   ω = 0.0653 rad/s

600 = 0.0653 x r

R = 9183 m

R = 9200 m (approx)

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.