## Q. 3.3

Finding an Acceleration Direction Geometrically

A moon orbits a planet in a circular path. At one instant, the moon’s velocity is $\overrightarrow{ v }_1$; at a later time, its velocity is $\overrightarrow{ v }_2$ (Fig. 3.14). The average acceleration is given by Equation 2.6:

$\overrightarrow{ a }_{ av , x} \equiv \frac{\Delta \overrightarrow{ v }_x}{\Delta t}=\frac{\Delta v_x}{\Delta t} \hat{ \imath }=\frac{v_x\left(t_f\right) – v_x\left(t_i\right)}{t_f-t_i} \hat{ \imath }=\frac{v_{f x}-v_{i x}}{t_f-t_i} \hat{ \imath }$          (2.6)

$\overrightarrow{ a }_{ av }=\frac{\Delta \vec{v}}{\Delta t}$

Find the direction of the average acceleration.

## Verified Solution

INTERPRET and ANTICIPATE
The time interval is a scalar. Therefore, the average acceleration must point in the same direction as the change in velocity. If we know the direction of $\Delta \vec{v}$, we also know the direction of $\vec{a}_{ av }$.

SOLVE
To determine the direction of $\Delta \vec{v}$, subtract  $\vec{v}_2-\vec{v}_1$ geometrically. We could use tail-to-tail subtraction, but we have arbitrarily chosen to multiply $\vec{v}_1$ by −1 and add the result to $\vec{v}_2$. The resultant vector $\Delta \vec{v}$ runs from the tail of $\vec{v}_2$ to the head of $– \vec{v}_1$. The average acceleration vector is proportional to the change in velocity and therefore points in the same direction.

CHECK and THINK
The two velocity vectors have the same length, so the moon’s speed is the same at both instants: $v_1 = v_2.$ It might seem surprising that we found a nonzero acceleration, but our answer is correct. In this case, the nonzero acceleration means that the direction of the moon’s velocity is changing $\overrightarrow{ v _1} \neq \overrightarrow{ v }_2$