2D, circular, and relative motion mini quiz

At the top, $v_{y}$ = 0 and $v_{x}$ remains 8 $\mathrm{m/s}$ .

$a_{x}$ = 0 and $a_{y}$ = −g throughout the ideal flight.

0 = 11.25− $1/2$ (10) $t^2$ gives $t^2$ = 2.25 and t = 1.5 s.

$\Delta x$ = $v_{x}$ t = 6.0(1.5) = 9.0 m.

Both components share the same time; $a_{x}$ = 0.

Its magnitude is fixed, but its tangent direction changes.

Changing the velocity vector requires acceleration toward the centre.

$a_{c}$ = $v^2/R$ .

$a_{c}$ = (3.0)²/1.5 = 6.0 $\mathrm{m/s^2}$ , directed toward the centre.

The required resultant is vertical, so the boat's horizontal component must cancel the current.

The required westward component is 1.5 $\mathrm{m/s}$ . The northward component is √(2.5²−1.5²) = 2.0 $\mathrm{m/s}$ .

sin θ = 1.5/2.5 gives θ ≈ 36.9° west of north.

When the horizontal components cancel, the speed relative to the ground is 2.0 $\mathrm{m/s}$ due north.

t = 120 m/(2.0 $\mathrm{m/s}$ ) = 60 s.

The figure marks the highest point of the trajectory at t = 2 s; the tangent is horizontal there.

Substituting t = 2 s gives r(2 s) = (4 m)i + (4 m)j.

$v(t)$ = (2.0 $\mathrm{m/s}$ )i + [(4.0 $\mathrm{m/s}$ ) − (2.0 $\mathrm{m/s^2}$ )t]j; therefore v(2 s) = (2.0 $\mathrm{m/s}$ )i.

$a(t)$ = −(2.0 $\mathrm{m/s^2}$ )j throughout the interval.

At that instant velocity is horizontal toward $+x$ , consistent with the trajectory's tangent, while acceleration points toward −y.

Add the motion of O relative to P to the motion of P relative to S.

This gives the velocity of O relative to S.

Velocity is tangent.

Radial acceleration points toward the centre, perpendicular to the instantaneous velocity.

t = √(2h/g).

t₄ₕ/tₕ = √(4h/h) = 2.

t = √[2(1.25)/10] = 0.50 s.

x = $v_{x}$ t = 4.0(0.50) = 2.0 m.

v_person/ground = v_person/walkway + v_walkway/ground.

v = 1.5 + 0.8 = +2.3 $\mathrm{m/s}$ .

T = 2π(2)/(4π) = 1 s.

$a_{c}$ = (4π)²/2 = 8π² ≈ 78.96 $\mathrm{m/s^2}$ .