distance from the initial to the final position of a point P. Thus, it is the length of an imaginary straight path, typically distinct from the path actually travelled by P. A displacement vector represents the length and direction of that imaginary straight path.
A position vector expresses the position of a point P in space in terms of a displacement from an arbitrary reference point O (typically the origin of a coordinate system). Namely, it indicates both the distance and direction of an imaginary motion along a straight line from the reference position to the actual position of the point.
A displacement may be also described as a relative position: the final position of a point () relative to its inBlogger: Membangun Indonesia dengan Fisika - Create postitial position (), and a displacement vector can be mathematically defined as the difference between the final and initial position vectors:
For motion over a given interval of time, the displacement divided by the length of the time interval defines the average velocity.
Distance versus directed distance and displacementnegative and distance travelled never decreases. Distance is a scalar quantity or a magnitude, whereas displacement is a vector quantity with both magnitude and direction.
The distance covered by a vehicle (for example as recorded by an odometer), person, animal, or object along a curved path from a point A to a point B should be distinguished from the straight line distance from A to B. For example whatever the distance covered during a round trip from A to B and back to A, the displacement is zero as start and end points coincide. In general the straight line distance does not equal distance travelled, except for journeys in a straight line.
VelocityIn physics, velocity is speed in a given direction. Speed describes only how fast an object is moving, whereas velocity gives both the speed and direction of the object's motion. To have a constant velocity, an object must have a constant speed and motion in a constant direction. Constant direction, typically constrains the object to motion in a straight path. A car moving at a constant 20 kilometers per hour in a circular path does not have a constant velocity. The rate of change in velocity is acceleration. Velocity is a vector physical quantity; both magnitude and direction are required to define it. The scalar absolute value (magnitude) of velocity is speed, a quantity that is measured in metres per second (m/s or ms−1) when using the SI (metric) system.
For example, "5 metres per second" is a scalar and not a vector, whereas "5 metres per second east" is a vector. The average velocity v of an object moving through a displacement during a time interval (Δt) is described by the formula:
The equations that apply to bodies moving linearly (in one dimension) with constant acceleration are often referred to as "SUVAT" equations where the five variables are represented by those letters (s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time); the five letters may be shown in a different order.
The body is considered between two instants in time: one initial point and one current (or final) point. Problems in kinematics may deal with more than two instants, and several applications of the equations are then required. If a is constant, the differential, a dt, may be integrated over an interval from 0 to Δt (Δt = t − ti), to obtain a linear relationship for velocity. Integration of the velocity yields a quadratic relationship for position at the end of the interval.
Relative velocityTo describe the motion of object A with respect to object B, when we know how each is moving with respect to a reference object O, we can use vector algebra. Choose an origin for reference, and let the positions of objects A, B, and O be denoted by rA, rB, and rO. Then the position of A relative to the reference object O is
For example, let Ann move with velocity relative to the reference (we drop the O subscript for convenience) and let Bob move with velocity , each velocity given with respect to the ground (point O). To find how fast Ann is moving relative to Bob (we call this velocity ), the equation above gives:
Motion graphs and derivativesIn mechanics, the derivative of the position vs. time graph of an object is equal to the velocity of the object. In the International System of Units, the position of the moving object is measured in meters relative to the origin, while the time is measured in seconds. Placing position on the y-axis and time on the x-axis, the slope of the curve is given by:
A similar fact also holds true for the velocity vs. time graph. The slope of a velocity vs. time graph is acceleration, this time, placing velocity on the y-axis and time on the x-axis. Again the slope of a line is change in y over change in x:
Since the velocity of the object is the derivative of the position graph, the area under the line in the velocity vs. time graph is the displacement of the object. (Velocity is on the y-axis and time on the x-axis. Multiplying the velocity by the time, the seconds cancel out and only meters remain. .)
The same multiplication rule holds true for acceleration vs. time graphs. When is multiplied by time (s), velocity is obtained. ().
Average acceleration is the change in velocity (Δv) divided by the change in time (Δt). Instantaneous acceleration is the acceleration at a specific point in time which is for a very short interval of time as Δt approaches zero.
Tangential and centripetal accelerationThe velocity of a particle moving on a curved path as a function of time can be written as:
Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet-Serret formulas.
Variable rates of changemean) rate of change is required. If the velocity or positions change non-linearly over time, such as in the example shown in the figure, then differentiation provides the correct solution. Differentiation reduces the time-spans used above to be extremely small and gives a velocity or acceleration at each point on the graph rather than between a start and end point. The derivative forms of the above equations are
integration is the opposite of differentiation, it is also possible to express position as a function of velocity and velocity as a function of acceleration. The process of determining the area under the curve, as described above, can give the displacement and change in velocity over particular time intervals by using definite integrals:
Ideal projectile motionIdeal projectile motion assumes that there is no air resistance. This assumption simplifies the math greatly, and is a close approximation of actual projectile motion in cases where the distances travelled are small. Ideal projectile motion is also a good introduction to the topic before adding the complications of air resistance.
- sin(x + y) = sin(x)cos(y) + sin(y)cos(x)
If x and y are same,
- sin(2x) = 2sin(x)cos(x)
Uneven GroundNow we will allow (y0) to be nonzero. Our equations of motion are now
The elevation angle which will provide the maximum range when launching the projectile from a non-zero initial height can be computed by finding the derivative of the range with respect to the elevation angle and setting the derivative to zero to find the extremum:
- where and horizontal range.
Substituting u = (cos θ)2 and 1 − u = (sin θ)2 produces:
Example: For the values g = 9.81m / s2, y0 = 40m , and v = 50m / s, an elevation angle θ = 41.1° produces a maximum range of Rmax = 292.1m.
Angle of impactThe angle ψ at which the projectile lands is given by:
Trajectory of a projectile
The United States Department of Defense and NATO define a ballistic trajectory as a trajectory traced after the propulsive force is terminated and the body is acted upon only by gravity and aerodynamic drag.
The following applies for ranges which are small compared to the size of the Earth.In the equations on this page, the following variables will be used:
- g: the gravitational acceleration—usually taken to be 9.81 m/s2 near the Earth's surface
- θ: the angle at which the projectile is launched
- v: the velocity at which the projectile is launched
- y0: the initial height of the projectile
- d: the total horizontal distance traveled by the projectile
Conditions at the final position of the projectile
[Distance travelledThe total horizontal distance (d) travelled.
Time of flightThe time of flight (t) is the time it takes for the projectile to finish its trajectory.
The above results are found in Range of a projectile.
Angle of reachThe "angle of reach" (not quite a scientific term) is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.
Conditions at an arbitrary distance x
Height at xThe height y of the projectile at distance x is given by
Velocity at xThe magnitude, | v | , of the velocity of the projectile at distance x is given by
DerivationThe magnitude |v| of the velocity is given by
Here the x-velocity remains constant; it is always equal to v cos θ.
The y-velocity can be found using the formula
- vf = vi + at
Angle θ required to hit coordinate (x,y)θ are:
First, two elementary formulae are called upon relating to projectile motion:
- (2b) (Trigonometric identity)
- (2c) (Trigonometric identity)
- (2d) (Algebra)
- (2e) (Substitution)
- (2f) (Quadratic formula)
- (2f) (Algebra)
- (2g) (Substitution)
- (2h) (Algebra)
Uphill/downhill in uniform gravity in a vacuum
Given a hill angle α and launch angle θ as before, it can be shown that the range along the hill Rs forms a ratio with the original range R along the imaginary horizontal, such that:
- (Equation 11)
While the same equation applies to projectiles fired uphill, the interpretation is more complex as sometimes the uphill range may be shorter or longer than the equivalent range along level terrain. Equation 11 may be set to Rs / R = 1 (i.e. the slant range is equal to the level terrain range) and solving for the "critical angle" θcr:
- "Rifleman's rule"
Derivation based on equations of a parabolaThe intersect of the projectile trajectory with a hill may most easily be derived using the trajectory in parabolic form in Cartesian coordinates (Equation 10) intersecting the hill of slope m in standard linear form at coordinates (x,y):
- (Equation 12) where in this case, y = dv, x = dh and b = 0
- (Solving above x)
Trajectory of a projectile with air resistance
For this assumption, that air resistance may be taken to be in direct proportion to the velocity of the particle is not correct for a typical projectile in air with a velocity above a few tens of meters/second, and so this equation should not be applied to that situation.
free body diagram on the right is for a projectile that experiences air resistance and the effects of gravity. Here, air resistance is assumed to be in the direction opposite of the projectile's velocity. Fair = − kv is written due to the initial assumption of direct proportionality implies that the air resistance and the velocity differ only by a constant arbitrary factor with units of N*s/m.
As an example, say that when the velocity of the projectile is 4 m/s, the air resistance is 7 newtons (N). When the velocity is doubled to 8 m/s, the air resistance doubles to 14 N accordingly. In this case, k = 7/4 N x s/m. Note that k is needed in order to relate the air resistance and the velocity by an equal sign: otherwise, it would be stating incorrectly that the two are always equal in value (i.e. 1 m/s of velocity gives 1 N of force, 2 m/s gives 2 N etc.) which isn't always the case, and also it keeps the equation dimensionally correct (a force and a velocity cannot be equal to each other, e.g. m/s = N). As another quick example, Hooke's Law (F = − kx) describes the force produced by a spring when stretched a distance x from its resting position, and is another example of a direct proportion: k in this case has units N/m (in metric).
To show why k = 7/4 N·s/m above, first equate 4 m/s and 7 N:
(Introduction of k)
For more on proportionality, see: Proportionality (mathematics)
The relationships that represent the motion of the particle are derived by Newton's Second Law, both in the x and y directions. In the x direction ΣF = − kvx = max and in the y direction ΣF = − kvy − mg = may.
This implies that: (1),
Solving (1) is an elementary differential equation, thus the steps leading to a unique solution for vx and, subsequently, x will not be enumerated. Given the initial conditions vx = vxo (where vxo is understood to be the x component of the initial velocity) and sx = 0 for t = 0:
While (1) is solved much in the same way, (2) is of distinct interest because of its non-homogeneous nature. Hence, we will be extensively solving (2). Note that in this case the initial conditions are used vy = vyo and sy = 0 when t = 0.
This first order, linear, non-homogeneous differential equation may be solved a number of ways, however, in this instance it will be quicker to approach the solution via an integrating factor: .
And by integration we find:
Solving for our initial conditions:
With a bit of algebra to simplify (3a):
An example is given using values for the mass and terminal velocity for a baseball taken from .
- m = 0.145 kg (5.1 oz)
- v0 = 44.7 m/s (100 mph)
- g = -9.81 m/s² (-32.2 ft/s²)
- vt = -33.0 m/s (-73.8 mph)
From Wikipedia, the free encyclopedia