What Is the Velocity Change of a Fighter Plane in Horizontal Circular Motion?

  • Thread starter Thread starter nesan
  • Start date Start date
  • Tags Tags
    Velocity
AI Thread Summary
A fighter plane moving in a horizontal circular path takes 9.8 seconds to change its heading by a quarter circle, with a radius of 2.0 x 10^3 meters. The velocity change can be determined by first calculating the angular velocity and then converting it to tangential speed. The correct approach involves using vector subtraction to find the difference between the initial and final velocities. The discussion emphasizes understanding the relationship between angular and linear motion, particularly through the formula ωr = v. Ultimately, the solution requires considering the arc distance and applying basic principles of circular motion.
nesan
Messages
74
Reaction score
0

Homework Statement



a fighter plane is moving in a horizontal circular path at a constant speed, it takes 9.8 seconds to change its heading by one quarter of a circle from north to east, if the radius of the curvature is 2.0 x 10^3, find

d) the velocity change

The Attempt at a Solution



v = d / t

d = square root ((2.0 x 10^3)^2 + (2.0 x 10^3)^2)

d ≈ 2828m

d / t = 288.6m/s

But the textbook says the answer is 4.5 x 10^2 m/s

I have no idea, in what other ways I can approach this. Help please. :( Pointing me in the right direction will do. ^_^ Thank you.
 
Physics news on Phys.org
The plane didn't take a straight line from the starting position to the final position over the 9.8 seconds.

Instead, why not think first of the angular velocity, since the plane is describing a circle and you're essentially given the angle it goes through over the 9.8 seconds. If you have the angular velocity and the radius, what is the corresponding tangential speed?

After that, what are the two vectors that represent the starting velocity and final velocity? What's the change between them?
 
gneill said:
The plane didn't take a straight line from the starting position to the final position over the 9.8 seconds.

Instead, why not think first of the angular velocity, since the plane is describing a circle and you're essentially given the angle it goes through over the 9.8 seconds. If you have the angular velocity and the radius, what is the corresponding tangential speed?

After that, what are the two vectors that represent the starting velocity and final velocity? What's the change between them?

Can you explain a bit more on what you mean by angular velocity please? Thank you. :)

EDIT : We haven't learned about tangential velocity and such, could you please explain it in-terms of a beginner, please and thank you. :)

NVM got it.

distance / time = speed

square root (speed ^2 + speed ^2) = average velocity

:)

Thank you. :)
 
Last edited:
How do you measure a distance if you're travel in a circle?
This must involve the arc distance along the circle.

Since the aircraft in flying in a circle, the angular speed is used.
But there's relation between angular and linear motion.
One of them velocity, ωr=v, where ω is angular speed dθ/dt(radian per second)

When you got the velocity at 2 points use vector subtraction to find the difference/change.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top