All categories

4 years ago

An airplane traveling from San Francisco northeast to Chicago travels 1260 km in 3.5 h. What is the airplanes average velocity?

Physics

1 answer:

IrinaK [193]4 years ago

6 0

Answer:

Average velocity = 360km/h due Northeast.

Explanation:

Given the following data;

Distance = 1260km

Time = 3.5h.

Velocity =?

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

$V = \frac{d}{t}$

Substituting into the above equation;

$V = \frac{1260}{3.5}$

Velocity, V = 360km/h.

Since velocity is a vector quantity, it must have both magnitude and direction.

<em>Hence, the average velocity of the airplane is 360km/h due Northeast. </em>

You might be interested in

The Rigidbody component adds collision to a GameObject

kifflom [539]

Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>

Explanation:

Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
You can indeed have a collider with no rigidbody. If there's no rigidbody then Unity assumes the object is static, non-moving.
If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.

7 0

3 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to