Kinematics

Kinematics is about motion of an object or particle. This page hopes to equip you with sufficient knowledge so you are not at a disadvantage compared to those also taking physics.

Contents

Key Concepts

The primary focus is motion in one direction, positive or negative. The particle may also move in other directions, but they do not impact this direction of interest. So for instance, there could be both xx and yy motion, but they can be examined separately. This is not some fundamental fact of physics, rather this is a single-variable calculus course and the problems are only so difficult.

The positive or negative signs in position, displacement, velocity, and acceleration denote the directions. In particular, quantities that are in the same direction have the same signs; ones in the opposite directions have opposite signs. The signs can be all different. For example, the position could be positive, velocity could be zero, at the same time acceleration is negative. This means the position and acceleration here are in opposite directions. When the correct signs are supplied to the calculus, the result will be encoded with the correct signs.

Almost always, time tt is non-negative. If tt were negative, it would represent what had happened prior to the start of the observation at t=0t = 0. The given equation is likely to be only valid for a specific interval in time. For example, gravity only impacts motion until the particle hits the ground.

Finally if you are taking physics, the acceleration there is constant. Because of the possibility of non-constant acceleration in math, do not rely on your physics kinematics formulas, which only work under constant acceleration.

Definitions

Displacement, ss, is the change in position on a number line, meaning both displacement and position can be positive or negative. When the displacement is with respect to the origin, the two are the same, ie x−0=xx - 0 = x. In other cases remember to add back the initial position to obtain final position.

s=x2−x1  ⟺  x2=s+x1s = x_2 - x_1 \iff x_2 = s + x_1

Because the origin can be freely moved, the easiest thing to do is to set the initial position to 0, ie x1=0x_1 = 0.

x1=0  ⟹  s=x2x_1 = 0 \implies s = x_2

Velocity vv, and acceleration aa are defined as

v(t)= ds dtv(t) = \frac{\d s}{\d t}
a(t)= dv dta(t) = \frac{\d v}{\d t}

Using integration, with velocity known at t1t_1 and position known at t2t_2, the equations become

s(t)=∫t1tv(τ) dτs(t) = \int_{t_1}^{t} v(\tau)\d \tau
v(t)=∫t2ta(τ) dτv(t) = \int_{t_2}^{t} a(\tau)\d \tau

where τ\tau is a variable of integration representing time, because we cannot use the same variable in the integrand and in the limits of integration.

distance

Distance is the absolute value of displacement.

d=∣s∣d = \lvert s\rvert

speed

Speed is the absolute value of velocity.

speed=∣v∣\text{speed} = \lvert v\rvert

distance traveled

Distance traveled is not distance. Instead, it is the length of path taken. It is

distance traveled=∫t1t2∣v(t)∣ dt\text{distance traveled} = \int_{t_1}^{t_2} \lvert v(t)\rvert \d t

When calculating this integral by hand, it typically involves

  1. solving v(t)=0v(t) = 0
  2. split v(t)v(t) into the regions separated by its zeros
  3. integrate for each region.
  4. add up the absolute values of the areas from step 3.

other formulas

a=v dv ds\displaystyle {a} = v \frac{\d v}{\d s}

This relies upon the chain rule.

Derivation:

a= dv dt= dv ds⋅ ds dt= dv ds⋅v\begin{align*} a &= \frac{\d v}{\d t} \\ &= \frac{\d v}{\d s} \cdot \frac{\d s}{\d t}\\ &= \frac{\d v}{\d s} \cdot v \\ \end{align*}

The integral version of this is

∫s1s2a ds=∫v1v2v dv=12(v22−v12)\int_{s_1}^{s_2} a\d s = \int_{v_1}^{v_2} v\d v = \frac 12 (v_2^2 - v_1^2)

Tips

  • Draw a diagram, this could be a sketch of the situation, but could also be a velocity-time diagram, as you could integrate to get displacement, or differentiate to get acceleration.
  • Signs matter within a question, but either forward or backward could be the positive direction. Ensure that displacement, velocity, and acceleration all have consistent (but necessarily the same) signs.
  • As mentioned, distance traveled is not distance.
  • a=v dv ds\displaystyle a = v\frac{\d v}{\d s} and its integral form are only useful when you know nothing about time.
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