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(11) Two planes approach cach other head-on. Each has aspeed of 780 $\mathrm{km} / \mathrm{h}$ , and they spot cach other when they areinitially 12.0 $\mathrm{km}$ apart. How much time do the pilots have totake evasive action?

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$27.7 \mathrm{s}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Rotation of Rigid Bodies

Dynamics of Rotational Motion

Equilibrium and Elasticity

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

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In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

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Our question says that two planes approach each other head on. Each has a speed of 780 kilometres per hour and they spot each other when there are initially 12 kilometers apart. How much time did the pilots have to take evasive action? Okay, so I wrote down what we were given in our problem and that was that. The velocity of plain one which I write his V one is equal to the velocity of plain to be too, and those air both traveling at 700 kilometers per hour and the distance between them when they spot each other is 12 kilometres. So now we need to figure out how much time they have before they collide. Well, they're total velocity since they're traveling, what towards one another ads, right. So the total velocity is the one plus fee to crank because of the time it's going to take for them. To meet head on would be the exact same as the time it would take for something travelling twice the speed of one of the objects to breach something that's at rest since they're both moving towards each other. So if you take the one and add it to be too. You end up with 1000 560 kilometers per hour. Okay, well, we know the velocity, and we know the distance. And since distance is equal to velocity, which ours is the total velocity times the time he confined the time cause we know velocity and distance. So the time is equal to the distance divided by the total velocity. Okay, so let's plug in those values on behalf. You have 12.0 kilometers for distance. We have 1560 kilometers per hour for speed. Okay. And Sophie, work this out. Well, the kilometer units are going to cancel on. You're gonna be left with this and units of ours. But it's going to be a fraction of an hour. And since fractions of an hour are very intuitive toe work with. Let's go. Ahem. Convert this hours that we have in two seconds. Well, we can do that by realizing there are 3000 600 seconds in every hour. So if we multiply this by 3600 over one, we will convert that from hours into seconds. And when you do that you realize that these plane pilots have 27 that's what this looks more like. A seven must like 2 27 0.7 seconds. So the time that they have is much shorter than the time it took for us to calculate this problem. So they should probably have some onboard computers that do this for them box it is their solution.

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