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 $\mathbf{10}\text{-}\mathbf{s}$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 interval, the astronaut is moving toward the right along the x-axis at $\mathbf{15}\mathbf{.}\mathbf{0}\mathit{m}\mathbf{/}\mathit{s}$, and at the end of the interval she is moving toward the right at$5.0m/s$ .

(b) At the beginning she is moving toward the left atrole="math" localid="1655276110547" $5.0m/s$ , and at the end she is moving toward the left at $15.0m/s$.

(c) At the beginning she is moving toward the right at , and at the end she is moving toward the left atrole="math" localid="1655276636193" $15.0m/s$ .

• The velocity changes are taking place at the 10-s interval.
• In a), the astronaut is moving towards the right of the x-axis at $15.0m/s$ in the beginning of the interval.
• In a), the astronaut is moving towards the right of the x-axis at $5.0m/s$in the end of the interval.
• In b), the astronaut is moving towards the left at $5.0m/s$ in the beginning of the interval.
• In b), the astronaut is moving towards the right of the x-axis at $15.0m/s$ in the end of the interval.
• In c), the astronaut is moving towards the right at $15.0m/s$ in the beginning of the interval.
• In c), the astronaut is moving towards the left at $15.0m/s$in the end of the interval.

The lawstates that the acceleration of an object is directly proportional to the rate of change of velocity and inversely proportional to the rate of change of time.

The rate of change of velocity divided by the rate of change of time gives the average acceleration of the astronaut.

From Newton’s second law, the average acceleration of the astronaut is expressed as:

$\begin{array}{rcl}a& =& \frac{\Delta v}{\Delta t}\\ & =& \frac{\Delta v_2-\Delta v_1}{\Delta t}\\ & & \end{array}$

Here, $\Delta v$is the rate of change of velocity and $\Delta t$is the rate of change of time.

a) The final velocity $\Delta v_2$is $5m/s$and the initial velocity$\Delta v_1$is $15m/s$and the time taken $\Delta t$is $10\text{s}$. Hence, substituting the values in the above equation, we get,

$\begin{array}{rcl}a& =& \frac{5m/s-15m/s}{10m/s}\\ & =& -1m/s^2\end{array}$

Thus, the acceleration is$-1m/s^2$and it is negative or towards the left.

b) the final velocity $\Delta v_2$is $15m/s$and the initial velocity$\Delta v_1$is $5m/s$and the time taken $\Delta t$is$10s$. Hence, substituting the values in the above equation, we get-

$\begin{array}{rcl}a& =& \frac{-15m/s-\left(-\right)5m/s}{}\\ & =& -1m/s^2\end{array}$

Thus, the acceleration is $-1m/s^2$and it is negative or towards the left.

c)the final velocity$\Delta v_2$ is $15m/s$and the initial velocity $\Delta v_1$is $15m/s$and the time taken $\Delta t$is$10s$ . Hence, substituting the values in the above equation, we get-

$\begin{array}{rcl}a& =& \frac{-15m/s-15m/s}{10s}\\ & =& -3m/s^2\end{array}$

Thus, the acceleration is$=-3m/s^2$ and it is negative or towards the left.