Velocity, displacement and force vectors

Displacement is the vector from a starting point to a finishing point. Distance travelled is the total length of the route. Speed is the magnitude of velocity; a force also has a magnitude and a direction.

A walk 4 km east, then 3 km north travels 7 km. Its displacement is 4i + 3j km, with magnitude 5 km. Returning to the starting point makes the total displacement zero while increasing the distance travelled.

In this lesson, i points east and j points north. State the units with every vector. A position vector in kilometres and a velocity in metres per second cannot be combined without converting units.

At velocity 8i − 6j km/h, the speed is 10 km/h. For 18 minutes, use t = 18/60 = 3/10 h.

Displacement = (3/10)(8i − 6j)
= (12/5)i − (9/5)j km
Distance = (3/10) × 10 = 3 km

For a velocity in cm/s and a time of 2 minutes, use 120 seconds. For two legs with different velocities, add their displacement vectors, but add their separate travelled distances as scalars.

A displacement of length d on bearing β has east component d sin β and north component d cos β:

A route of 6 km on bearing 150°, then 4 km on bearing 030°, has displacements 3i − 3√3 j and 2i + 2√3 j. The resultant is 5i − √3 j km, of magnitude 2√7 km.

The resultant is southeast, so its bearing is 90° + tan⁻¹(√3/5) ≈ 109.1°. Keep the exact components until this final angle calculation. The total walked distance is 10 km.

The bearing from the endpoint back to the start is 180° opposite, here about 289.1°. For a general non-zero displacement, use its east/north signs to choose the correct quadrant; the zero displacement has no bearing.

For the model boat, solve r₀ + tv = b. To reach B(5,−2), the equations are:

−3 + 2t = 5 ⇒ t = 4
2 − t = −2 ⇒ t = 4

Both agree and t is positive, so the boat reaches B after 4 hours. Its velocity is a positive multiple of the displacement from its start to B.

A boat’s velocity relative to the water is 5i km/h, while the water’s velocity relative to the bank is −i + 2j km/h. The boat’s velocity relative to the bank is their vector sum:

v = 5i + (−i + 2j) = 4i + 2j km/h
Speed relative to bank = √20 = 2√5 km/h

Add vectors before taking a magnitude. Adding the two speeds, 5 + √5, would not account for their different directions. The model assumes the current and the boat’s velocity through the water stay constant.

A particle changes velocity from i + 4j m/s to 7i − 2j m/s over 3 seconds. Its average acceleration is:

a = [(7i − 2j) − (i + 4j)]/3
= 2i − 2j m/s²
|a| = 2√2 m/s²

A particle can change velocity by changing direction even when its speed is unchanged. Conversely, knowing two endpoint velocities gives an average acceleration, not proof that the instantaneous acceleration was constant.

If F₁ = 2i − j N and F₂ = pi + qj N, the resultant is R = (p + 2)i + (q − 1)j N. Requiring a non-zero resultant parallel to i + 2j gives:

R = λ(i + 2j), λ ≠ 0
p + 2 = λ,   q − 1 = 2λ
q = 2p + 5, with p ≠ −2

If p = 1, then q = 7 and R = 3i + 6j N, with magnitude 3√5 N. A negative λ would give the opposite direction. A bare parallel condition does not choose between those two directions.

Constant velocity means a straight path with unchanged speed and direction. It may approximate a short interval of drifting or steady travel. It is a poor long-term model for a kicked ball that slows, changes direction, bounces or leaves the ground.

A particle model ignores the object’s size and rotation. A constant-current model ignores changing flow. Flat-map east/north coordinates ignore curvature over large journeys. State the assumptions relevant to the calculation, then explain how breaking one could change the answer.

Forces are vectors too, but a general rigid-body mechanics problem can depend on where they act. The simple resultant-force questions here concern a particle or a supplied translational model.