Dynamics, Kinematics, and Vibrations

A. Kinematics of particles

Particle Kinematics: Constant Acceleration and Projectiles

Describe how a particle moves — rectilinear and curvilinear paths, the constant-acceleration kinematic equations, projectiles, normal/tangential and polar components, and relative motion.

Kinematics is the grammar of dynamics: it describes how a particle moves — position, velocity, acceleration — without ever asking what force is responsible. Almost every Dynamics question on the FE Mechanical exam begins here, because before you can write

\sum F = ma

you must know the acceleration, and before you can use work-energy you must relate speed to distance. Master the four constant-acceleration equations and the handful of component frames (Cartesian, normal-tangential, polar) and a large fraction of the topic becomes bookkeeping. The FE Reference Handbook's Dynamics chapter ('Particle Kinematics') lays out exactly these relations; this concept turns them into reflexes.

Velocity and acceleration as derivatives

Position is a vector

\mathbf{r}(t)

; velocity is its time derivative and acceleration is the derivative of velocity. In one dimension this is the chain that ties the three together, and the third form —

a\,ds = v\,dv

— is the one students forget, yet it is the bridge to work-energy because it eliminates time. Use it whenever a problem gives you a relationship between

a

and position

s

but no clock.

v = \dfrac{ds}{dt}, \qquad a = \dfrac{dv}{dt} = v\,\dfrac{dv}{ds}

The constant-acceleration equations

When acceleration is constant, the calculus collapses into three algebraic relations plus a time-free fourth. These are the workhorses: free fall (with

a = g = 9.81\ \text{m/s}^2

or

32.2\ \text{ft/s}^2

downward), braking cars, launched masses. Pick the equation that contains your three knowns and one unknown — if the problem never mentions time, reach immediately for the

v^2

form so you never have to solve for

t

first.

\begin{aligned} v &= v_0 + a t \\ s &= s_0 + v_0 t + \tfrac{1}{2} a t^2 \\ s &= s_0 + \tfrac{1}{2}(v_0 + v) t \\ v^2 &= v_0^2 + 2 a (s - s_0) \end{aligned}

Projectile motion

A projectile is just two constant-acceleration problems running in parallel: horizontal motion with

a_x = 0

(constant velocity) and vertical motion with

a_y = -g

. The two are linked only through the shared time

t

. The launch angle

\theta

splits the initial speed into

v_{0x} = v_0\cos\theta

and

v_{0y} = v_0\sin\theta

; solve the vertical equation for the time of an event, then feed that time into the horizontal equation for range. There is no horizontal force, so horizontal velocity never changes — this is the single most common conceptual win on these problems.

x = v_0\cos\theta\,t, \qquad y = v_0\sin\theta\,t - \tfrac{1}{2} g t^2

Normal and tangential components

On a curved path it is natural to resolve acceleration along the direction of motion (tangential) and toward the center of curvature (normal). The tangential component

a_t = dv/dt

changes the speed; the normal component

a_n = v^2/\rho

changes the direction and always points inward. Even at constant speed (

a_t = 0

) a particle on a curve accelerates, because

a_n

is nonzero — this is why a car rounding a bend needs friction. The total magnitude is the Pythagorean sum of the two.

a_t = \dfrac{dv}{dt}, \qquad a_n = \dfrac{v^2}{\rho}, \qquad a = \sqrt{a_t^2 + a_n^2}

Radial and transverse (polar) components

When motion is described by a radius

r

and angle

\theta

— a particle on a rotating arm, a cam follower — polar components are cleanest. Note the cross terms: the radial acceleration carries a

-r\dot\theta^2

(the centripetal term), and the transverse acceleration carries the

2\dot r\dot\theta

Coriolis term that appears whenever the particle moves radially while the arm rotates. Dropping that Coriolis term is a classic error on rotating-arm problems.

a_r = \ddot r - r\dot\theta^2, \qquad a_\theta = r\ddot\theta + 2\dot r\dot\theta

Relative motion

When two particles move, the position, velocity, and acceleration of

A

relative to a translating frame on

B

add vectorially:

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

. Treat each as a vector — break into components, add, then recombine. Relative-velocity problems (a boat in a current, two cars at an intersection) are pure vector addition; the trap is mixing up which body is the reference, so always read the subscript order as 'of A with respect to B.'

\mathbf{r}_A = \mathbf{r}_B + \mathbf{r}_{A/B}, \quad \mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}, \quad \mathbf{a}_A = \mathbf{a}_B + \mathbf{a}_{A/B}

Exam strategy

First decide the frame: straight line

\to

constant-acceleration set; curve at known speed

\to

normal-tangential; rotating arm

\to

polar; two bodies

\to

relative-motion vectors. For projectiles, write the

x

and

y

equations separately and link them through

t

; the time to peak is

v_{0y}/g

and total flight (level ground) is twice that. If time is not asked and not given, use

v^2 = v_0^2 + 2a\,\Delta s

to skip it entirely. Keep

a_n = v^2/\rho

in mind even when speed is constant — the exam loves the 'constant speed therefore zero acceleration' trap, which is false on any curve.

Key equations

Kinematic derivatives

v = \dfrac{ds}{dt}, \quad a = \dfrac{dv}{dt} = v\,\dfrac{dv}{ds}

Defines velocity and acceleration. The third form eliminates time; use it for acceleration given as a function of position $s$ (m), $v$ in m/s, $a$ in m/s².

Constant-acceleration velocity

v = v_0 + a t

Velocity after time $t$ under constant $a$ ; $v_0$ = initial velocity (m/s), $a$ in m/s², $t$ in s.

Constant-acceleration position

s = s_0 + v_0 t + \tfrac{1}{2} a t^2

Displacement under constant acceleration; $s_0$ = initial position (m).

Average-velocity displacement

s = s_0 + \tfrac{1}{2}(v_0 + v) t

Displacement as average speed times time; valid only for constant $a$ . Useful when $a$ is not wanted.

Time-free (velocity-position)

v^2 = v_0^2 + 2 a (s - s_0)

Relates speed and distance with no time term — use when $t$ is neither given nor sought.

Projectile equations

x = v_0\cos\theta\,t, \quad y = v_0\sin\theta\,t - \tfrac{1}{2} g t^2

Horizontal $a_x=0$ , vertical $a_y=-g$ ; $\theta$ = launch angle above horizontal, $g = 9.81\ \text{m/s}^2$ .

Normal/tangential acceleration

a_t = \dfrac{dv}{dt}, \quad a_n = \dfrac{v^2}{\rho}, \quad a = \sqrt{a_t^2 + a_n^2}

$a_t$ changes speed, $a_n$ (toward center) changes direction; $\rho$ = radius of curvature (m).

Polar (radial/transverse) acceleration

a_r = \ddot r - r\dot\theta^2, \quad a_\theta = r\ddot\theta + 2\dot r\dot\theta

$2\dot r\dot\theta$ is the Coriolis term; $r$ in m, $\dot\theta$ in rad/s, $\ddot\theta$ in rad/s².

Plane circular motion

v = r\omega, \quad a_n = r\omega^2 = \dfrac{v^2}{r}, \quad a_t = r\alpha

Constant-radius case: $\omega$ = angular velocity (rad/s), $\alpha$ = angular acceleration (rad/s²).

Relative velocity

\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}

Vector addition; $\mathbf{v}_{A/B}$ = velocity of $A$ as seen from a frame translating with $B$ .

Worked examples

Projectile range and peak height

Problem. A ball is launched from ground level at $v_0 = 25\ \text{m/s}$ at $40^\circ$ above horizontal. Find the time of flight, horizontal range, and maximum height. Take $g = 9.81\ \text{m/s}^2$ .

Solution. Split the launch velocity:

v_{0x} = 25\cos 40^\circ = 19.15\ \text{m/s}

,

v_{0y} = 25\sin 40^\circ = 16.07\ \text{m/s}

. Time of flight (return to

y=0

):

t = 2 v_{0y}/g = 2(16.07)/9.81 = 3.28\ \text{s}

. Range:

R = v_{0x}\,t = 19.15 \times 3.276 = 62.7\ \text{m}

. Peak height (where

v_y = 0

):

h = v_{0y}^2/(2g) = 16.07^2/(2\cdot 9.81) = 13.2\ \text{m}

. Answers:

t = 3.28\ \text{s}

,

R = 62.7\ \text{m}

,

h = 13.2\ \text{m}

. Sanity: range formula

R = v_0^2\sin 2\theta/g = 625\sin 80^\circ/9.81 = 62.7\ \text{m}

agrees.

R = \dfrac{v_0^2 \sin 2\theta}{g} = \dfrac{25^2 \sin 80^\circ}{9.81} = 62.7\ \text{m}

Accelerating car: final speed and time

Problem. A car moving at $15\ \text{m/s}$ accelerates uniformly at $2.5\ \text{m/s}^2$ over a distance of $80\ \text{m}$ . Find the final speed and the elapsed time.

Solution. No time is given, so use the time-free equation first:

v^2 = v_0^2 + 2a\,\Delta s = 15^2 + 2(2.5)(80) = 225 + 400 = 625

, so

v = 25.0\ \text{m/s}

. Then

t = (v - v_0)/a = (25 - 15)/2.5 = 4.00\ \text{s}

. Answers:

v = 25.0\ \text{m/s}

,

t = 4.00\ \text{s}

. Sanity: average speed

(15+25)/2 = 20\ \text{m/s}

over

4\ \text{s}

gives

80\ \text{m}

— consistent.

Total acceleration on a curve

Problem. A car travels at $20\ \text{m/s}$ on a circular track of radius $150\ \text{m}$ while speeding up at $3.0\ \text{m/s}^2$ . Find the magnitude of its total acceleration.

Solution. Normal (centripetal) component:

a_n = v^2/\rho = 20^2/150 = 2.67\ \text{m/s}^2

, directed toward the center. Tangential component:

a_t = 3.0\ \text{m/s}^2

, along the direction of motion. Total:

a = \sqrt{a_n^2 + a_t^2} = \sqrt{2.67^2 + 3.0^2} = \sqrt{7.11 + 9.0} = 4.01\ \text{m/s}^2

. Answer:

a = 4.01\ \text{m/s}^2

. Sanity: even at 'constant 20 m/s' the car would still have

a_n = 2.67\ \text{m/s}^2

, so total acceleration is never zero on a curve.

Common pitfalls

•Assuming constant speed means zero acceleration. On any curve $a_n = v^2/\rho \ne 0$ even when $a_t = 0$ — direction is changing.
•Forgetting horizontal velocity is constant in projectiles. There is no horizontal force, so $v_x = v_0\cos\theta$ throughout; never apply gravity to the $x$ equation.
•Mixing degrees and radians. Angular formulas ( $v = r\omega$ , $a_n = r\omega^2$ ) require $\omega$ in rad/s; convert rpm with $\omega = 2\pi N/60$ .
•Dropping the Coriolis term $2\dot r\dot\theta$ in $a_\theta$ when a particle slides along a rotating arm. It is present whenever both $\dot r$ and $\dot\theta$ are nonzero.
•Using the $v^2 = v_0^2 + 2a\Delta s$ form with non-constant acceleration. The constant-acceleration set is only valid when $a$ truly does not change; otherwise integrate.
•Reversing relative-motion subscripts. $\mathbf{v}_{A/B}$ is 'A relative to B'; $\mathbf{v}_A = \mathbf{v}_B + \mathbf{v}_{A/B}$ , not $\mathbf{v}_B - \mathbf{v}_{A/B}$ .
•Sign errors on vertical projectile motion. Pick a sign convention (up positive) and apply $a_y = -g$ consistently to both velocity and position.

References

NCEES FE Reference Handbook — Dynamics
Hibbeler, Engineering Mechanics: Dynamics — particle kinematics chapters

C. Newton’s second law for particles

Particle Kinetics: Newton’s Second Law, Work–Energy and Impulse–Momentum

Relate forces to motion three ways — Newton’s second law with friction, the work–energy theorem with conservative forces, and linear impulse–momentum with impact.