## Chapter 3

## 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.

## Step-by-Step

## 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