Average Acceleration and Instantaneous Acceleration — Question 6

Question 6

A constant acceleration need not point along displacement

A particle begins at the origin with $\vec v_0=(4,1)$ m/s and constant $\vec a=(-2,3)$ m/s$^2$.

1. Find v and r at 2 s.

2. Find the time when the horizontal velocity vanishes and decide whether the particle stops.

3. Find the average acceleration and average velocity over the first 2 s. Explain why these vectors need not be parallel.

Question 6 - Solution

A constant acceleration need not point along displacement

(a) Apply the component equations.

$$\vec v(2)=(4,1)+2(-2,3)=\boxed{(0,7)\,\mathrm{m/s}},\quad\vec r(2)=2(4,1)+2(-2,3)=(4,8)\,\mathrm m$$ $\vec v=\vec v_0+\vec a t$ and $\vec r=\vec v_0t+\tfrac12\vec a t^2$. At 2 s these give $\vec v=(0,7)$ m/s and $\vec r=(4,8)$ m. The two vector equations carry different powers of time.

(b) Check the other component at the same time.

$$4-2t=0\ \Rightarrow\ t=2\,\mathrm s,\quad v_y=1+3(2)=7\,\mathrm{m/s}\ne0$$ $v_x=(4-2t/\mathrm s)\ \mathrm{m/s}$ vanishes at 2 s, but $v_y=7$ m/s then. The motion has a turning x coordinate without coming to rest. The x acceleration remains negative after this instant, making x motion reverse.

(c) Average different physical changes.

$$\vec a_{\mathrm{av}}=\frac{(0,7)-(4,1)}2=(-2,3)\,\mathrm{m/s^2},\quad\vec v_{\mathrm{av}}=(2,4)\,\mathrm{m/s}$$ Average acceleration is $(-2,3)$ m/s$^2$, equal to the constant a. Average velocity is $(4,8)\ \mathrm m/(2\ \mathrm s)=(2,4)$ m/s. Acceleration tracks change of velocity; average velocity tracks change of position, so their directions are not constrained to agree.