Question

2.16 • an astronaut has left the international space station to test a new space scooter. her partner measures the following velocity changes, each taking place in a 10 s interval. what are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? assume that the positive direction is to the right. (a) at the beginning of the inter - val, the astronaut is moving toward the right along the x - axis at 15.0 m/s, and at the end of the interval she is moving toward the right at 5.0 m/s. (b) at the beginning she is moving toward the left at 5.0 m/s, and at the end she is moving toward the left at 15.0 m/s. (c) at the beginning she is moving toward the right at 15.0 m/s, and at the end she is moving toward the left at 15.0 m/s.

Explanation:

Step1: Recall average - acceleration formula

The formula for average acceleration is $a_{avg}=\frac{\Delta v}{\Delta t}=\frac{v_f - v_i}{\Delta t}$, where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $\Delta t$ is the time - interval. Here, $\Delta t = 10\ s$.

Step2: Solve part (a)

Given $v_i = 15.0\ m/s$ and $v_f = 5.0\ m/s$. Then $a_{avg}=\frac{v_f - v_i}{\Delta t}=\frac{5.0 - 15.0}{10}=\frac{- 10.0}{10}=-1.0\ m/s^{2}$. The magnitude of the average acceleration is $|a_{avg}| = 1.0\ m/s^{2}$, the algebraic sign is negative, and the direction is to the left.

Step3: Solve part (b)

Given $v_i=-5.0\ m/s$ (negative because moving to the left) and $v_f=-15.0\ m/s$. Then $a_{avg}=\frac{v_f - v_i}{\Delta t}=\frac{-15.0-(-5.0)}{10}=\frac{-15.0 + 5.0}{10}=\frac{-10.0}{10}=-1.0\ m/s^{2}$. The magnitude of the average acceleration is $|a_{avg}| = 1.0\ m/s^{2}$, the algebraic sign is negative, and the direction is to the left.

Step4: Solve part (c)

Given $v_i = 15.0\ m/s$ and $v_f=-15.0\ m/s$. Then $a_{avg}=\frac{v_f - v_i}{\Delta t}=\frac{-15.0 - 15.0}{10}=\frac{-30.0}{10}=-3.0\ m/s^{2}$. The magnitude of the average acceleration is $|a_{avg}| = 3.0\ m/s^{2}$, the algebraic sign is negative, and the direction is to the left.

Answer:

(a) Magnitude: $1.0\ m/s^{2}$, Sign: negative, Direction: left
(b) Magnitude: $1.0\ m/s^{2}$, Sign: negative, Direction: left
(c) Magnitude: $3.0\ m/s^{2}$, Sign: negative, Direction: left