Timedora

3 - Two-Dimensional Kinematics

A practical guide to analyzing two-dimensional motion with vectors, component equations, projectile models, and relative velocity.

Representing Position and

describes motion in a plane, such as the motion of a thrown object, a swimmer crossing a river, or a vehicle following a curved path. The central idea is to represent physical quantities as vectors and analyze perpendicular components separately.

A reliable solution begins by choosing a coordinate system. A common choice is a positive horizontal xx-axis, a positive upward yy-axis, and an origin at a convenient location. Position can then be written as

r⃗=xi^+yj^.\vec r=x\hat i+y\hat j.

The between an initial position and a final position is

Δr⃗=r⃗f−r⃗i.\Delta\vec r=\vec r_f-\vec r_i.

In component form,

Δr⃗=(xf−xi)i^+(yf−yi)j^.\Delta\vec r=(x_f-x_i)\hat i+(y_f-y_i)\hat j.

Because has both magnitude and direction, it is not the same as total distance traveled. Its magnitude is

∣Δr⃗∣=(Δx)2+(Δy)2,\lvert\Delta\vec r\rvert=\sqrt{(\Delta x)^2+(\Delta y)^2},

and its direction can be found from tan⁡ϕ=ΔyΔx\tan\phi=\frac{\Delta y}{\Delta x}, with the signs of the components used to select the correct quadrant.

Takeaway: Establish axes first, preserve component signs, and distinguish the vector from its magnitude.

Resolving and Adding Vectors

A vector can be resolved into perpendicular components before equations are applied. If a vector has magnitude AA and points at an angle θ\theta above the positive xx-axis, then

Ax=Acos⁡θ,Ay=Asin⁡θ.A_x=A\cos\theta,\qquad A_y=A\sin\theta.

The vector is reconstructed as

A⃗=Axi^+Ayj^.\vec A=A_x\hat i+A_y\hat j.

Given components, recover the magnitude and direction with

A=Ax2+Ay2,θ=tan⁡−1(AyAx).A=\sqrt{A_x^2+A_y^2}, \qquad \theta=\tan^{-1}\left(\frac{A_y}{A_x}\right).

The inverse tangent provides a reference angle, so inspect the signs of AxA_x and AyA_y to determine the actual direction.

Add vectors component by component:

A⃗+B⃗=(Ax+Bx)i^+(Ay+By)j^.\vec A+\vec B=(A_x+B_x)\hat i+(A_y+B_y)\hat j.

For example, if A⃗=(3i^+4j^) m\vec A=(3\hat i+4\hat j)\,\text{m} and B⃗=(−1i^+2j^) m\vec B=(-1\hat i+2\hat j)\,\text{m}, then

A⃗+B⃗=(2i^+6j^) m,\vec A+\vec B=(2\hat i+6\hat j)\,\text{m},

with magnitude

∣A⃗+B⃗∣=22+62=6.3 m.\lvert\vec A+\vec B\rvert=\sqrt{2^2+6^2}=6.3\,\text{m}.

Do not add vector magnitudes unless the vectors point in the same direction.

Takeaway: Resolve, combine, and then recombine components; scalar magnitudes alone do not preserve direction.

Component Kinematics with Constant Acceleration

For average motion, divide the vector change by the elapsed time. Average velocity is

v⃗avg=Δr⃗Δt,\vec v_{\text{avg}}=\frac{\Delta\vec r}{\Delta t},

so its components are vx,avg=ΔxΔtv_{x,\text{avg}}=\frac{\Delta x}{\Delta t} and vy,avg=ΔyΔtv_{y,\text{avg}}=\frac{\Delta y}{\Delta t}. Average acceleration is

a⃗avg=Δv⃗Δt,\vec a_{\text{avg}}=\frac{\Delta\vec v}{\Delta t},

with component relationships ax,avg=ΔvxΔta_{x,\text{avg}}=\frac{\Delta v_x}{\Delta t} and ay,avg=ΔvyΔta_{y,\text{avg}}=\frac{\Delta v_y}{\Delta t}.

For constant acceleration, use a separate one-dimensional equation for each axis:

x=x0+v0xt+12axt2,vx=v0x+axt,vx2=v0x2+2ax(x−x0),x=x_0+v_{0x}t+\frac{1}{2}a_xt^2, \qquad v_x=v_{0x}+a_xt, \qquad v_x^2=v_{0x}^2+2a_x(x-x_0),

and

y=y0+v0yt+12ayt2,vy=v0y+ayt,vy2=v0y2+2ay(y−y0).y=y_0+v_{0y}t+\frac{1}{2}a_yt^2, \qquad v_y=v_{0y}+a_yt, \qquad v_y^2=v_{0y}^2+2a_y(y-y_0).

The same time tt must be used in both directions because the components describe simultaneous parts of one motion.

Motion graphs provide equivalent information: the slope of a position-versus-time graph is velocity, the slope of a velocity-versus-time graph is acceleration, the area under a velocity-versus-time graph is , and the area under an acceleration-versus-time graph is change in velocity.

Takeaway: Treat the axes as separate calculations, but connect them with shared time and shared initial conditions.

For , after launch the only significant force is gravity, with air resistance neglected. Near Earth's surface,

a⃗=0i^−gj^,ax=0,ay=−g,\vec a=0\hat i-g\hat j, \qquad a_x=0, \qquad a_y=-g,

where g≈9.8 m/s2g\approx 9.8\,\text{m/s}^2. If the launch speed is v0v_0 at an angle θ\theta above the horizontal, resolve the initial velocity as

v0x=v0cos⁡θ,v0y=v0sin⁡θ.v_{0x}=v_0\cos\theta, \qquad v_{0y}=v_0\sin\theta.

The component equations are

x=x0+(v0cos⁡θ)t,x=x_0+(v_0\cos\theta)t,
y=y0+(v0sin⁡θ)t−12gt2,y=y_0+(v_0\sin\theta)t-\frac{1}{2}gt^2,
vx=v0cos⁡θ,vy=v0sin⁡θ−gt.v_x=v_0\cos\theta, \qquad v_y=v_0\sin\theta-gt.

Thus, horizontal velocity remains constant, while vertical velocity changes continuously. At the highest point, vy=0v_y=0, but generally vx≠0v_x\ne 0, and the acceleration remains downward rather than becoming zero.

When launch and landing heights are equal, useful specialized results are

tflight=2v0sin⁡θg,t_{\text{flight}}=\frac{2v_0\sin\theta}{g},
Δymax⁡=(v0sin⁡θ)22g,\Delta y_{\max}=\frac{(v_0\sin\theta)^2}{2g},

and

R=v02sin⁡(2θ)g.R=\frac{v_0^2\sin(2\theta)}{g}.

These formulas require equal launch and landing heights and negligible air resistance. For unequal heights, use the general component equations instead.

A horizontal-launch example illustrates the method. A ball leaves a table at 4.0 m/s4.0\,\text{m/s} from a height of 1.25 m1.25\,\text{m}. Taking the launch point as y0=0y_0=0 and the floor as y=−1.25 my=-1.25\,\text{m}, use

−1.25=−12(9.8)t2-1.25=-\frac{1}{2}(9.8)t^2

to obtain t=0.505 st=0.505\,\text{s}. The horizontal distance is

Δx=(4.0)(0.505)=2.02 m,\Delta x=(4.0)(0.505)=2.02\,\text{m},

so the ball lands approximately 2.0 m2.0\,\text{m} from the table's edge.

Takeaway: Gravity affects the vertical component, not the horizontal component, in the ideal projectile model.

Relative Motion and Reference Frames

Velocity is always measured relative to a . An object can be stationary relative to one frame and moving relative to another. For objects AA and BB, the relative position and velocity are

r⃗A/B=r⃗A−r⃗B,\vec r_{A/B}=\vec r_A-\vec r_B,
v⃗A/B=v⃗A−v⃗B.\vec v_{A/B}=\vec v_A-\vec v_B.

Equivalently,

v⃗A=v⃗A/B+v⃗B.\vec v_A=\vec v_{A/B}+\vec v_B.

For a car moving east at 20 m/s20\,\text{m/s} and a truck moving east at 12 m/s12\,\text{m/s},

v⃗car/truck=20i^−12i^=8i^ m/s.\vec v_{\text{car/truck}}=20\hat i-12\hat i=8\hat i\,\text{m/s}.

If the truck instead moves west at 12 m/s12\,\text{m/s}, its velocity is v⃗truck=−12i^ m/s\vec v_{\text{truck}}=-12\hat i\,\text{m/s}, so

v⃗car/truck=20i^−(−12i^)=32i^ m/s.\vec v_{\text{car/truck}}=20\hat i-(-12\hat i)=32\hat i\,\text{m/s}.

For a swimmer moving at 2.0 m/s2.0\,\text{m/s} north relative to still water while the river flows east at 1.0 m/s1.0\,\text{m/s}, the ground velocity is

v⃗swimmer/ground=(1.0i^+2.0j^) m/s.\vec v_{\text{swimmer/ground}}=(1.0\hat i+2.0\hat j)\,\text{m/s}.

Its ground speed is

v=(1.0)2+(2.0)2=2.24 m/s.v=\sqrt{(1.0)^2+(2.0)^2}=2.24\,\text{m/s}.

The swimmer travels downstream while crossing. To move directly north, the swimmer must aim partly west so that the westward swimming component cancels the river's eastward component.

Takeaway: Relative-motion equations are vector equations. Subtract signed components, not speeds alone.

A General Problem-Solving Strategy

Use this sequence for most problems.

  1. Define the system and coordinates. Identify the object, , origin, and positive directions.

  2. Draw the motion. Include axes, positions, velocity vectors, acceleration vectors, and relevant angles.

  3. Resolve vectors. Express every known vector in xx- and yy-components, keeping signs consistent.

  4. Write component equations. Use one-dimensional kinematics independently along each axis. Do not substitute a scalar speed for a velocity component.

  5. Use shared variables. The same time interval connects the horizontal and vertical equations.

  6. Recombine when needed. Find a resultant speed or direction from

v=vx2+vy2,θ=tan⁡−1(vyvx).v=\sqrt{v_x^2+v_y^2}, \qquad \theta=\tan^{-1}\left(\frac{v_y}{v_x}\right).

  1. Check the result. Confirm that units are consistent, signs match the diagram, a zero acceleration component produces constant velocity in that direction, and the final direction lies in the correct quadrant.

Frequent errors include confusing distance with , treating speed as a vector, assigning gravity a horizontal component in ideal , setting acceleration to zero at the top of a trajectory, applying the equal-height range formula to unequal heights, and adding magnitudes instead of components.

Final takeaway: A complete solution combines a clear coordinate system, signed , separate axis equations, shared time, appropriate units, and a physical reasonableness check.