Kinematics mini quiz

$x=-20\,m$ locates the object on the negative side of the origin.

It does not tell us how x changes or how v changes.

The student travels 10 m outward and 6 m on the return: distance = 16 m.

$\Delta x=4\,m-0\,m=+4\,m$ .

v and a both point toward $-x$ .

When they have the same direction, the magnitude $|v|$ increases.

$v=\frac{-4-2}{3-0}=\frac{-6}{3}=-2\,m/s$ .

Position decreases by two metres per second relative to the chosen axis.

v = 4.0 + (−1.0)(3.0) = 1.0 $\mathrm{m/s}$ .

$\Delta x$ = 4.0(3.0)+ $1/2$ (−1.0)(3.0)² = 7.5 m.

$0=12-3t$ gives t = 4 s.

$v(6)=12-18=-6\,m/s$ : it is already moving toward $-x$ .

The vertical component changes from positive to negative and is zero at the turning instant.

Gravity continues to act: $a_{y}$ = −g throughout the ideal flight.

The graph rises from 0 m to 4 m between 0 s and 2 s, remains horizontal until 5 s, and falls to −2 m at 7 s.

The ratios $\Delta x/\Delta t$ are +2 $\mathrm{m/s}$ , 0 $\mathrm{m/s}$ , and −3 $\mathrm{m/s}$ , respectively.

Position does not change between 2 s and 5 s; this is the rest interval.

Displacement is −2 m − 0 m = −2 m. Distance is 4 m + 0 m + 6 m = 10 m.

The successive slopes of the polyline are positive, zero, and negative. The curve crosses v = 0 at t = 7 s.

The slopes give +2 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , −2 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .

The signed areas are 4 m, 12 m, 4 m, and −2 m; therefore $\Delta x$ = 18 m.

Distance adds the absolute values of the areas: 4 m + 12 m + 4 m + 2 m = 22 m.

After the start, direction changes when the curve crosses v = 0 at t = 7 s.

The line crosses v = 0 at t = 2 s: maximum height occurs there.

The positive triangular area is $1/2$ (2 s)(20 $\mathrm{m/s}$ ) = 20 m.

Between 2 s and 4 s there is a triangle with area −20 m; the total signed area is 0 m.

Distance adds the magnitudes of both areas: 20 m + 20 m = 40 m.

Speed increases linearly, remains constant, and then decreases linearly to zero.

The accelerations are +1 $\mathrm{m/s^2}$ , 0 $\mathrm{m/s^2}$ , and −1 $\mathrm{m/s^2}$ .

The two triangles contribute 2 m each, and the central rectangle contributes 12 m.

Total displacement is 2 m + 12 m + 2 m = 16 m, positive in the chosen direction.

Between 0 s and 2 s, $\Delta v$ = (+2 $\mathrm{m/s^2}$ )(2 s) = +4 $\mathrm{m/s}$ ; thus v(2 s) = 3 $\mathrm{m/s}$ .

Since a = 0 between 2 s and 5 s, v(5 s) = 3 $\mathrm{m/s}$ .

Between 5 s and 9 s, $\Delta v$ = (−1 $\mathrm{m/s^2}$ )(4 s) = −4 $\mathrm{m/s}$ ; therefore v(9 s) = −1 $\mathrm{m/s}$ .

In the first segment, −1 + 2t = 0 gives t = 0.5 s. In the last, 3 − (t−5) = 0 gives t = 8 s.

The two crossings separate motion toward $-x$ , then $+x$ , and finally $-x$ .

The final displacement is zero.

v̄ = $\Delta x/\Delta t$ = 0.

If $\vec a=d\vec v/dt$ = 0, v does not change.

A constant velocity need not be zero.

The figure starts at x = 0 m and ends at x = −2 m.

$\Delta x$ = −2 − 0 = −2 m.

$\Delta x$ = +20 m − 2 m = +18 m.

$x_{f}$ = $x_{i}$ + $\Delta x$ = −3 m + 18 m = 15 m.

The areas are 2 m, 12 m, and 2 m.

Total displacement is 16 m.

0 = 12² + 2(−3) $\Delta x$ .

$\Delta x$ = 24 m.

d = |17 − 2| + |8 − 17| = 24 m.

$\Delta x$ = 8 − 2 = +6 m.

a = (24 − 0)/8.0.

a = 3.0 $\mathrm{m/s^2}$ .

$\Delta x$ ₁ = 18 m and $\Delta x$ ₂ = −8 m.

$\Delta x$ = 10 m; $\Delta t$ = 10 s.

v̄ = 10/10 = 1 $\mathrm{m/s}$ .

Moving toward $-x$ implies v < 0.

As the magnitude of a negative velocity decreases, a points toward $+x$ : a > 0.