Acceleration and uniformly accelerated motion

In a chosen reference frame the position of a body is given by the radius vector $\vec r(t)$, while the displacement $\Delta\vec r=\vec r_2-\vec r_1$ depends only on the initial and final points, whereas the path $s$ is merely a scalar measure of the trajectory travelled. The average velocity is $\vec v_{avg}=\Delta\vec r/\Delta t$, and the instantaneous velocity $\vec v=d\vec r/dt$ is tangent to the trajectory. Acceleration is the rate of change of the velocity vector: $\vec a=\frac{\Delta\vec v}{\Delta t}\quad(\Delta t\to 0).$ The vector $\vec a$ is not tied to the direction of motion: it may point along $\vec v$ (speeding up), opposite to $\vec v$ (slowing down), or at an angle to $\vec v$ (bending the trajectory). In uniformly accelerated motion $\vec a=\text{const}$, hence $\vec v=\vec v_0+\vec a t,\qquad \Delta\vec r=\vec v_0 t+\tfrac{1}{2}\vec a t^2,\qquad v^2=v_0^2+2\vec a\cdot\Delta\vec r.$ The first two formulas are vector relations, while the third is a scalar relation (linking speeds and the projection of acceleration onto displacement); in 2D and 3D the first two hold independently for each coordinate, which allows the motion to be split along the axes of the chosen reference frame. Because the path equals the magnitude of the displacement only when there is no reversal, the sign of the velocity component must be checked separately.

A cyclist moves in a straight line along the positive $x$-axis. In $3.0$ s the speed increases from $4.0$ m/s to $10.0$ m/s. Find the acceleration vector and the displacement during this time.

A car travels along a straight road at $20$ m/s. The driver brakes, and the acceleration is constant at $5.0$ m/s² in magnitude, directed opposite to the motion. How long does the car take to stop, and what distance does it cover before stopping? Compare the directions of $\vec v$ and $\vec a$.

A drone flies with uniform acceleration in the $xy$-plane. At time $t_1$ its velocity is $\vec v_1=(4.0\,\hat i+2.0\,\hat j)$ m/s, and $\Delta t=2.0$ s later it is $\vec v_2=(2.0\,\hat i+6.0\,\hat j)$ m/s. Find the acceleration vector, its magnitude and angle with the $x$-axis, and the drone's displacement during these $2.0$ s.

Self-check in a kinematics simulation (e.g. PhET “The Moving Man” or “Projectile Motion”).

1. Set $v_0=0$ and a constant $a>0$; sketch the graphs $v(t)$ (a straight line with slope $a$) and $x(t)$ (a parabola). Note that the slope of the $v(t)$ graph equals the acceleration.
2. Change the sign of the acceleration so that $\vec a$ is opposite to $\vec v$: verify that the body slows down, stops momentarily, and then moves back. Record the moment when the velocity component changes sign.
3. Switch to 2D mode and check two cases:
   • If $\vec a\perp\vec v_0$ (e.g. a horizontal launch), initially the speed does not change in magnitude, but the direction curves; for $t>0$ the speed increases as $\sqrt{v_0^2+a^2t^2}$.
   • If $\vec a\cdot\vec v_0<0$ (e.g. a launch upward at an angle), the speed first decreases and the trajectory curves; split the motion along the $x$ and $y$ axes.

Before each run write down a prediction (the sign of $a_x$, the shape of the trajectory), then compare it with what you observe. Do all observations in the simulation; if you perform an experiment with an inclined plane at home, use only low speeds, an even floor, and do not throw objects.

Show that in uniformly accelerated motion the acceleration vector is constant and governs how the velocity vector changes, and link this vector picture to the $v(t)$ and $x(t)$ graphs.

A sketch: an $x$-axis with the body drawn at equally spaced instants; at each position the velocity vectors are drawn growing in length, together with one and the same acceleration vector $\vec a$, identical in length and direction, attached to the body. Beside it, two matching graphs: $v(t)$ as a straight line with slope $a$ and $x(t)$ as a parabola, where the sign of the slope of $v(t)$ matches the sign of the projection of $\vec a$ onto the $x$-axis.

A sprinter starts from rest and reaches $10$ m/s in $2.5$ s, then runs almost uniformly. The acceleration during the start is $a = \dfrac{10 - 0}{2.5} = 4$ m/s$^2$; the distance covered while accelerating is $s = \dfrac{v^2 - v_0^2}{2a} = \dfrac{100}{8} = 12.5$ m.

This is a direct check of the kinematic formulas $v = v_0 + at$ and $s = v_0 t + \dfrac{at^2}{2}$ for motion along one axis: the acceleration vector points the same way as the velocity vector, so the speed grows by $4$ m/s every second.

A car travels at $90$ km/h $= 25$ m/s and brakes to a full stop in $4$ s. Its acceleration is $a = \dfrac{0 - 25}{4} = -6.25$ m/s$^2$, and the braking distance is $s = \dfrac{25^2}{2 \cdot 6.25} = 50$ m — about fifty metres, the length of a swimming pool.

Here the acceleration is directed against the velocity vector, and the vector expression $\vec a = \dfrac{\Delta \vec v}{\Delta t}$ gives a negative projection onto the axis of motion: the speed decreases linearly while the distance grows as the square of time, exactly as $s = v_0 t - \dfrac{|a| t^2}{2}$ predicts.

A stone is released from rest from a bridge $20$ m high; air resistance is neglected, $g = 9.8$ m/s$^2$. The fall time is $t = \sqrt{\dfrac{2h}{g}} \approx 2.0$ s, and the speed at the water surface is $v = gt \approx 20$ m/s $\approx 72$ km/h.

Free fall is uniformly accelerated motion with a constant acceleration vector $\vec a = \vec g$ pointing downward: each second the velocity vector grows by $9.8$ m/s, and the distances covered in $1$ s, $2$ s, $3$ s are in the ratio $1 : 4 : 9$.