Motion and Forces flashcards

Distance is the total path length traveled (a scalar, never negative). Displacement is the straight-line change from start to finish with a direction (a vector). One full lap: 400 m distance, zero displacement.

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t}$, the change in velocity (m/s) divided by time (s), in ${\text{m/s}}^2$. From 2 to 8 m/s in 3 s gives $2{\text{m/s}}^2$.

$v=v_0+at$ and $\Delta x=v_{0}t+\frac{1}{2}a{t}^2$, with $v_0$ the initial velocity (m/s), a the acceleration (${\text{m/s}}^2$) and t the time (s). From rest at $4{\text{m/s}}^2$ for 3 s: 12 m/s and 18 m.

$F_{net}=ma$: net force in newtons equals mass in kg times acceleration in ${\text{m/s}}^2$. Subtract opposing forces first: 40 N forward minus 16 N friction on 8 kg gives $3{\text{m/s}}^2$.

No. Constant velocity (even at high speed) means zero acceleration, so the net force is zero; forces are balanced. A net force is needed only to change speed or direction (Newton's first law, inertia).

They are equal and opposite but act on two different objects. Only forces on the same object can cancel. A book's weight and the table's normal force balance, but they are not a third-law pair.

$W=mg$, with m in kg and g about $9.8{\text{m/s}}^2$ on Earth (often rounded to 10), giving newtons. Mass never changes with location; a 12 kg toolbox weighs about 118 N on Earth and 19 N on the Moon.

Equal distances at different speeds take unequal times, so use total distance over total time. 6 km out at 12 km/h (0.5 h) and back at 4 km/h (1.5 h) averages 12 km in 2 h = 6 km/h, not 8.