91Ó°ÊÓ

• 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:

$\begin{array}{rcl}a& =& \frac{\mathrm{Δ}v}{\mathrm{Δ}t}\\ & =& \frac{\mathrm{Δ}{v}_2-\mathrm{Δ}{v}_1}{\mathrm{Δ}t}\\ & & \end{array}$

Here, $\mathrm{Δ}v$is the rate of change of velocity and $\mathrm{Δ}t$is the rate of change of time.

a) The final velocity $\mathrm{Δ}{v}_2$is $5m/s$and the initial velocity$\mathrm{Δ}{v}_1$is $15m/s$and the time taken $\mathrm{Δ}t$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 $\mathrm{Δ}{v}_2$is $15m/s$and the initial velocity$\mathrm{Δ}{v}_1$is $5m/s$and the time taken $\mathrm{Δ}t$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$\mathrm{Δ}{v}_2$ is $15m/s$and the initial velocity $\mathrm{Δ}{v}_1$is $15m/s$and the time taken $\mathrm{Δ}t$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.