Motion in a Straight Line

A vector has magnitude and direction. Displacement (the straight-line change in position) has a direction, so it is a vector. Distance, speed, time and mass have size only, so they are scalars.

Displacement is measured in a straight line from start to finish. The runner ends exactly where they began, so the displacement is $0\text{ m}$. (The distance travelled is $400\text{ m}$, which shows the two are not the same.)

Average speed is distance over time: $v = \dfrac{d}{t} = \dfrac{150}{30} = 5.0\text{ m/s}$.

Velocity is a vector, so it changes if either the speed or the direction changes. Going around a bend the direction is changing, so the velocity is changing even though the speed (a scalar) stays at $60\text{ km/h}$.

Taking east as positive, $\Delta s = (+8) + (-3) = +5\text{ m}$, i.e. $5\text{ m}$ east. (The distance walked is $8 + 3 = 11\text{ m}$, but displacement is measured start-to-finish.)

$a = \dfrac{v - u}{t} = \dfrac{30 - 10}{5.0} = 4.0\text{ m/s}^2$.

$a = \dfrac{v - u}{t} = \dfrac{5.0 - 25}{4.0} = -5.0\text{ m/s}^2$. The negative sign shows the acceleration opposes the motion, i.e. the train is decelerating.

Average velocity uses displacement, not distance. Displacement $= 300 - 100 = 200\text{ m north}$ over a total time of $120\text{ s}$: $v_{av} = \dfrac{200}{120} \approx 1.7\text{ m/s north}$. (The average speed would be $\frac{400}{120} = 3.3\text{ m/s}$.)

On a displacement-time graph the velocity is the gradient. In the first $4\text{ s}$ the line rises from $0$ to $20\text{ m}$: $v = \dfrac{\Delta x}{\Delta t} = \dfrac{20}{4} = 5.0\text{ m/s}$. (After $4\text{ s}$ the line is flat, so the object is at rest.)

Split the area into a triangle and a rectangle. Triangle ($0$ to $3\text{ s}$): $\tfrac{1}{2}(3)(12) = 18\text{ m}$. Rectangle ($3$ to $5\text{ s}$): $(2)(12) = 24\text{ m}$. Total $= 18 + 24 = 42\text{ m}$.

At the top the velocity is momentarily zero, but gravity never switches off: the acceleration is still $9.8\text{ m/s}^2$ downward the whole time. Zero velocity does not mean zero acceleration, which is exactly why the ball does not stay at the top.

Using $v = u + at$ with $u = 0$: $v = 0 + (3.0)(8.0) = 24\text{ m/s}$.

Using $s = ut + \tfrac{1}{2}at^2$ with $u = 0$: $s = 0 + \tfrac{1}{2}(3.0)(8.0)^2 = \tfrac{1}{2}(3.0)(64) = 96\text{ m}$.

Use the equation without time, with $v = 0$ and $a = -5.0\text{ m/s}^2$: $v^2 = u^2 + 2as \Rightarrow 0 = 20^2 + 2(-5.0)s$, so $s = \dfrac{400}{10} = 40\text{ m}$.

In free fall $v = u + gt$ with $u = 0$: $v = (9.8)(2.0) = 19.6\text{ m/s}$.

At the top the velocity is momentarily zero. Taking up as positive, $v = u - gt \Rightarrow 0 = 19.6 - 9.8t$, so $t = \dfrac{19.6}{9.8} = 2.0\text{ s}$.

Average speed is total distance over total time, not the average of the speeds. Time out $= \frac{60}{60} = 1\text{ h}$; time back $= \frac{60}{40} = 1.5\text{ h}$; total $2.5\text{ h}$ for $120\text{ km}$: $v_{av} = \dfrac{120}{2.5} = 48\text{ km/h}$. More time is spent going slowly, so the average is below $50\text{ km/h}$.

Speed is the magnitude of the gradient. From $0$ to $4\text{ s}$: $\frac{20}{4} = 5\text{ m/s}$. From $6$ to $8\text{ s}$ the line drops from $20$ to $0\text{ m}$: $\frac{-20}{2} = -10\text{ m/s}$, a speed of $10\text{ m/s}$. The steepest section is $6$ to $8\text{ s}$ (the object is moving fastest, though backward).

By symmetry the time up is half the total flight: $2.0\text{ s}$. From the top, the ball falls from rest for $2.0\text{ s}$: $s = \tfrac{1}{2}gt^2 = \tfrac{1}{2}(9.8)(2.0)^2 = 19.6\text{ m}$. (Using the full $4.0\text{ s}$ would give four times too much.)

Total displacement is the area under the graph. Triangle ($0$–$8\text{ s}$): $\tfrac{1}{2}(8)(20) = 80\text{ m}$. Rectangle ($8$–$20\text{ s}$): $(12)(20) = 240\text{ m}$. Triangle ($20$–$25\text{ s}$): $\tfrac{1}{2}(5)(20) = 50\text{ m}$. Total $= 370\text{ m}$, so $v_{av} = \dfrac{370}{25} = 14.8\text{ m/s}$.

Time is unknown, so use $v^2 = u^2 + 2as$ with $u = 0$: $30^2 = 0 + 2a(90)$, giving $900 = 180a$ and $a = 5.0\text{ m/s}^2$.

Take up as positive; the ground is $20\text{ m}$ below the launch point, so $-20 = 15t - 4.9t^2$, giving $4.9t^2 - 15t - 20 = 0$. The quadratic formula gives $t = \dfrac{15 + \sqrt{225 + 392}}{9.8} \approx 4.1\text{ s}$ (the positive root). The ball first rises for $1.5\text{ s}$, then falls all the way to the base. ($2.0\text{ s}$ is the trap of treating it as simply dropped.)

Taking up as positive: $19.6 = 24.5t - 4.9t^2$, so $4.9t^2 - 24.5t + 19.6 = 0$, i.e. $t^2 - 5t + 4 = 0$ and $(t-1)(t-4) = 0$. The ball passes that height twice: at $t = 1\text{ s}$ going up and $t = 4\text{ s}$ coming down.

During the reaction time the car moves at constant speed: $20 \times 0.75 = 15\text{ m}$. Braking distance: $v^2 = u^2 + 2as \Rightarrow s = \dfrac{20^2}{2(5.0)} = 40\text{ m}$. Total $= 15 + 40 = 55\text{ m}$.

Take down as positive with $u = 5.0\text{ m/s}$: $s = ut + \tfrac{1}{2}gt^2 = (5.0)(2.0) + \tfrac{1}{2}(9.8)(2.0)^2 = 10 + 19.6 = 29.6\text{ m}$. The initial downward throw adds the $10\text{ m}$ on top of the free-fall drop.