Kinematics

A curve of a force F, vs displacement x (F vs x), represents the magnitude of a force as it is producing a displacement of a body. The area under the curve from

a point x1, to point x2, represents the work done by the force;

W =

⌠Fdx

If the force is constant from x1 to x2, then; W =

F∙(x

2 - x1)

The slope of the curve at a given value of x, (dF

/dx),

tells us how the force F is

varying with displacement x at that point.

For the case of a constant force, the value of the slope is zero, (dF

/dx

=

0),

meaning that the force is not varying as the displacement takes place.

Twice the speed of sound. That's the definition of mach number.

Mach number is a more useful measurement of speed than miles per hour in some ways (especially if you're designing an airplane), but not so good for answering questions like "So when are we getting to Denver?" since the speed of sound varies depending on the density of the air.

the actual speed of mach 2 is 2472 km/h which is twice mach 1 (1236km/h) at this speed the airplane is roughly 2 meters shorter. planes that can travel this fast are;

• e111-a raven
• mig 25 foxbat
• f22- raptor
• f22- strike raptor
• yf 12a blackbird
• concorde
• f 35 c

Velocity is a vector quantity and so has an associated direction.

Speed is a scalar quantity and hence does not.
Velocity is a vector and speed is a scalar Velocity is different because it also includes it's magnitude and direction. If positive is left and negative is right. Say a car is going in the positive direction then decides to stop and back up at the same speed it was going before. When it is backing up, it will be going in a negative direction. However, if it was speed, it's speed wouldn't change. Speed does not measure the magntidude or direction.
Velocity is speed with direction . The formula of velocity is :

v= delta "r" / t [ velocity = movement / time ]

Speed is a movement without direction .The formula of speed is :

v= S / t

Imagine a car moving forward for 50meters and then moving backwards for 20 meters .

The movement in velocity (delta r) --> 50-20= 30m

The "s" in Speed ---> 50+20= 70m

so in velocity you gotta take into consideration , whether the car moves forward or backwards . As far as speed is concerned , the only thing you need to do is summing up all the distance that the car's made .

Firing from a hill Firing a projectile from an elevated position increases its range. If you know the initial velocity, you should be able to use the usual formulas for displacement (distance) in the horizontal and vertical directions to determine the initial vertical position.1 When you say the initial velocity is known, I assume that includes magnitude and direction. Since velocity is a vector, you should be able to calculate the vertical and horizontal components.2 If you know the horizontal velocity and the horizontal displacement (distance traveled), you should be able to calculate the time in flight. Once you determine the time in flight, you should be able to use that value in the formula for vertical displacement to determine the initial vertical displacement. Hint: The vertical displacement of the projectile when it hits the ground is zero (assuming you have selected the origin -- the axes of the plane in which the projectile is moving -- properly). ----------- 1. d = d0 + V0t + [1/2]at2, where d0 is the initial displacement, v0 is the initial velocity, and a is acceleration. For motion in the vertical direction, a = -g. For motion in the horizontal direction, a = 0 (for projectile problems). 2. Vx = Vcos(theta); Vy = Vsin(theta), where theta is the angle of elevation. Maximum range is achieved when theta = 45 degrees. At that angle, Vx = Vy.

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PLEASE NOTE: formatting has been messed up in this so things that are supposed to be raised to a power, the number is not a superscript. This needs to be relooked at.

Maximum range is achieved when theta=45o only if the vertical displacement is zero (i.e. the projectile begins and ends at the same elevation). If launched from a certain height h, the angle for maximum range is given by

anglemax = 1/2 cos-1 [(gh)/(v^2 + gh)]

Returning to the problem, let h = launch height, R = horizontal distance from base of launch site to landing spot, V = launch speed, A = launch angle and T = time in air. The horizontal component of the launch velocity is constant since there is no acceleration in that direction. Therefore:

Vx = R/T

V cosA = R/T

Solving for T:

T = R/[V cosA]

Consider the vertical part of the problem. This solution is given for a projectile launched from an angle above the horizontal so that the initial vertical component of the velocity is positive when the acceleration due to gravity (g) is negative. Also assumed is that the launch position is above the landing position. Let the initial position be the origin.

d = do + viT + 1/2aT^2

-h = (V sinA)T - 1/2gT^2

Substituting for T:

(equation A) -h = [VR sinA ]/[V cosA] - [gR^2]/[2V^2 cos2A]

(equation B) h = -R [tanA] + [gR^2]/[2V^2cos2A]

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If you want to find the launch angle for a given height and launch speed that gives the maximum range, multiply both sides of equation A by 2cos2(theta) and rearrange to get:

(g/v^2)R^2 - (2sinAcosA)R - 2hcos2A=0

Using these trig identities:

2sinAcosA = sin2A

cos2A = 1/2 [1 + cos2A]

the equation becomes:

(g/v^2)R^2 - (sin2A)R - h[1+cos2A] = 0

Solving for R using the quadratic formula:

R(A) = v^2/(2g)[sin2A + (sin22A + (8gh/v^2)cos2A)1/2]

Find the derivative of

R(A): R'(A)=v^2/(2g)[2cos2A+1/2(sin22A+(8gh/v^2)cos2A)-1/2(4sin2Acos2A+(8gh/v^2)(-2sinAcosA))]

Set this equal to zero to find angle (A) for maximum range (R).

Rearrange and use some trig identities to get:

1/(cos22A) - v^2/(gh)(1/cos2A) - (1+v^2/(gh))=0

Use the quadratic formula to solve for 1/(cos2A):

1/(cos2A) = (1/2)[v^2/(gh) + (v^4/(g^2h^2)+4(1+v^2/(gh)))1/2] 1/(cos2A) = v^2/(gh) + 1

A = 1/2cos-1[(gh)/(v^2+gh)]