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Definition/Summary
Three simple equations of motion may be used when acceleration is constant in both magnitude and direction, or when acceleration is constant in magnitude and the motion is forced to follow a fixed track.
These three equations apply only in the direction of the acceleration: in perpendicular directions, the component of acceleration is zero, and so the component of displacement (distance) will be a constant times time.
Each equation involves only four of the five variables a\ u\ v\ s\ \text{and}\ t (other symbols used are u_i\ \text{and}\ u_f for u\ \text{and}\ v, and \Delta s\ \text{and}\ \Delta t for s\ \text{and}\ t), so choose whichever equation omits the variable you are not interested in.
s in these equations is the displacement (distance) relative to a stationary point. If the displacement given is between two bodies of which one has constant acceleration and the other has constant velocity, the equations may be applied in a frame in which the latter is stationary.
Equations
In the direction of constant acceleration:
v\ =\ u\ +\ at
v^2\ =\ u^2\ +\ 2as
s\ =\ ut\ +\ \frac{1}{2}at^2
Perpendicular to the direction of constant acceleration:
v\ =\ u
s\ =\ ut
For circular motion, with angular displacement \theta, angular velocity \omega, and angular acceleration \alpha:
\omega_f\ =\ \omega_i\ +\ \alpha t
\omega_f ^2\ =\ \omega_i^2\ +\ 2\alpha\theta
\theta\ =\ \omega_it\ +\ \frac{1}{2}\alpha t^2
Extended explanation
Motion of a projectile:
A ball thrown at speed V at an angle \theta above the horizontal will, using the usual (x,y) coordinates, have acceleration (0,-g), and horizontal and vertical components of displacement:
Why two solutions?
Exam questions often ask you to find the speed angle or time for a ball to reach its target.
There are usually two correct solutions, one is a "drive" and the other is a "lob", but they both take the same time, t, to get to the same place, (x,y).
They result from simultaneously solving the above equations, giving a quadratic equation in t.
Relative motion:
If body A has constant acceleration, and body B has constant velocity, and if the initial distance between them is given (or sought), then change to a frame of reference in which B is stationary.
The acceleration in that frame will be the same, but A's velocity will be replaced by its velocity relative to B.
In that frame, the given distance is now measured relative to a stationary point, and may be used as s.
Motion along a curved track:
A vehicle constantly accelerating or braking along a curved track obeys the same three equations.
Circular motion:
For an object moving in a circle, the same three equations may be used with angular acceleration velocity and displacement instead of ordinary (linear) acceleration velocity and displacement.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
Three simple equations of motion may be used when acceleration is constant in both magnitude and direction, or when acceleration is constant in magnitude and the motion is forced to follow a fixed track.
These three equations apply only in the direction of the acceleration: in perpendicular directions, the component of acceleration is zero, and so the component of displacement (distance) will be a constant times time.
Each equation involves only four of the five variables a\ u\ v\ s\ \text{and}\ t (other symbols used are u_i\ \text{and}\ u_f for u\ \text{and}\ v, and \Delta s\ \text{and}\ \Delta t for s\ \text{and}\ t), so choose whichever equation omits the variable you are not interested in.
s in these equations is the displacement (distance) relative to a stationary point. If the displacement given is between two bodies of which one has constant acceleration and the other has constant velocity, the equations may be applied in a frame in which the latter is stationary.
Equations
In the direction of constant acceleration:
v\ =\ u\ +\ at
v^2\ =\ u^2\ +\ 2as
s\ =\ ut\ +\ \frac{1}{2}at^2
Perpendicular to the direction of constant acceleration:
v\ =\ u
s\ =\ ut
For circular motion, with angular displacement \theta, angular velocity \omega, and angular acceleration \alpha:
\omega_f\ =\ \omega_i\ +\ \alpha t
\omega_f ^2\ =\ \omega_i^2\ +\ 2\alpha\theta
\theta\ =\ \omega_it\ +\ \frac{1}{2}\alpha t^2
Extended explanation
Motion of a projectile:
A ball thrown at speed V at an angle \theta above the horizontal will, using the usual (x,y) coordinates, have acceleration (0,-g), and horizontal and vertical components of displacement:
x\ =\ (V\cos\theta )t
y\ =\ (V\sin\theta )t\ -\ \frac{1}{2}gt^2
y\ =\ (V\sin\theta )t\ -\ \frac{1}{2}gt^2
Why two solutions?
Exam questions often ask you to find the speed angle or time for a ball to reach its target.
There are usually two correct solutions, one is a "drive" and the other is a "lob", but they both take the same time, t, to get to the same place, (x,y).
They result from simultaneously solving the above equations, giving a quadratic equation in t.
Relative motion:
If body A has constant acceleration, and body B has constant velocity, and if the initial distance between them is given (or sought), then change to a frame of reference in which B is stationary.
The acceleration in that frame will be the same, but A's velocity will be replaced by its velocity relative to B.
In that frame, the given distance is now measured relative to a stationary point, and may be used as s.
Motion along a curved track:
A vehicle constantly accelerating or braking along a curved track obeys the same three equations.
Circular motion:
For an object moving in a circle, the same three equations may be used with angular acceleration velocity and displacement instead of ordinary (linear) acceleration velocity and displacement.
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!