Physics · Ch 3 — Motion in a Plane

Average and Instantaneous Acceleration

3.3.2

Average and Instantaneous Acceleration

The definitions of acceleration in two dimensions mirror those of rectilinear motion exactly, with each quantity now carrying x and y components. The average acceleration of a particle between times $t_1$ and $t_2$ , with velocities $\vec{v}_1$ and $\vec{v}_2$ , is $\vec{a}_{av} = \dfrac{\vec{v}_2 - \vec{v}_1}{t_2-t_1} = (a_{av})_x\,\hat{i} + (a_{av})_y\,\hat{j}$ , where $(a_{av})_x = \dfrac{(v_2)_x - (v_1)_x}{t_2-t_1}$ and $(a_{av})_y = \dfrac{(v_2)_y - (v_1)_y}{t_2-t_1}$ . Its magnitude and direction are $a_{av} = \sqrt{(a_{av})_x^2 + (a_{av})_y^2}$ and $\tan\theta = \dfrac{(a_{av})_y}{(a_{av})_x}$ . …

Misc Ex.3.4Velocity and acceleration of three particles from their position vectors

Worked out. Three particles have position vectors given as functions of time t (the first constant/independent of t, the second linear in t with equal x and y rates of 5 m/s each, the third with a constant x-component of 5 m/s but a y-component growing as 10t^2, i.e. y-velocity 20t). The problem asks for the velocity and acceleration of each particle in SI units. The method differentiates each position vector once (dr/dt) to get velocity and twice (d^2r/dt^2) to get acceleration component-wise, then combines the x and y components using the Pythagorean magnitude formula and tan(theta) = (y-component)/(x-component) for direction: particle 1 is at rest (zero velocity); particle 2 moves with constant velocity 5i + 5j m/s (magnitude 5√2 m/s at 45° to the horizontal); particle 3 has a growing y-velocity 20t j m/s combined with a constant x-velocity 5 m …

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