Average Acceleration and Instantaneous Acceleration — Question 1

Question 1

A turn produces acceleration without a speed change

A vehicle has velocity $(6,0)$ m/s at the beginning of a 3 s interval and $(0,6)$ m/s at its end. Average acceleration is $\vec a_{\mathrm{av}}=\Delta\vec v/\Delta t$. No details of the intervening turn are given.

1. Find average acceleration, its magnitude and direction.

2. Compare the initial and final speeds. Why does their difference not determine average acceleration?

3. Can these data establish constant acceleration or constant speed throughout the interval?

Question 1 - Solution

A turn produces acceleration without a speed change

(a) Subtract the complete velocity vectors.

$$\vec a_{\mathrm{av}}=\frac{(0,6)-(6,0)}3=(-2,2)\,\mathrm{m/s^2},\quad\boxed{|\vec a_{\mathrm{av}}|=2\sqrt2\,\mathrm{m/s^2},\ \theta=135^\circ}$$ $\Delta\vec v=(-6,6)$ m/s, so $\vec a_{\mathrm{av}}=(-2,2)$ m/s$^2$. Its magnitude is $2\sqrt2\approx2.828$ m/s$^2$, directed $135^\circ$ from positive x. It points along a change in velocity, not along either endpoint velocity.

(b) Direction is part of velocity.

$$|\vec v_f|-|\vec v_i|=6-6=0,\quad|\vec v_f-\vec v_i|=6\sqrt2\,\mathrm{m/s}\ne0$$ Both endpoint speeds are 6 m/s, so the net speed change is zero. Their velocity vectors nevertheless differ. Subtracting magnitudes discards the directional change responsible for the nonzero average acceleration.

(c) Endpoint rates do not fix an intermediate history.

$$\vec v_{\mathrm{linear}}(t)=(6-2t,2t)\,\mathrm{m/s},\quad v_{\mathrm{linear}}(1.5)=3\sqrt2\,\mathrm{m/s}<6\,\mathrm{m/s}$$ Neither claim follows. Many time-dependent velocity histories share these endpoints. A constant acceleration equal to the average is one possible history, but its speed would not remain constant. A constant-speed turn is another possibility with changing acceleration direction.