91Ó°ÊÓ

Velocity in time interval $t=0$to$t=5.00\min$is zero

Velocity in time interval $t=5.00\min$to$t=10.0\min$ is$2.20\mathrm{m}/\mathrm{s}$ .

The ratio of the displacement to the time interval in which the displacement occurs is known as average velocity.

The formula to find the average velocity is given as follows,

$V_{avg}=\frac{∆x}{∆t}=\frac{x_2-x_1}{t_2-t_1}$

(i)

Here, $x_1$is the position at time $t_1$and$x_2$is the position at time $t_2$.

The ratio of the change in velocity over a time interval to that time interval is termed as average acceleration.

The formula to find the average acceleration is given as follows,

$a_{avg}=\frac{∆v}{∆t}=\frac{v_2-v_1}{t_2-t_1}$

(ii)

Here, $v_1$is the velocity at time $t_1$and $v_2$is the velocity at time $t_2$.

Initially, man is at the origin. The total time interval is,

$$\begin{gathered} ∆\mathrm{t}=8-2 \\ =6\min \\ =6×60\mathrm{s} \\ =360\mathrm{s} \end{gathered}$$

Sub-interval in which man is moving,

$$\begin{gathered} ∆{\mathrm{t}}^{\text{'}}=8-5 \\ =3\min \\ =3×60\mathrm{s} \\ =180\mathrm{s} \end{gathered}$$

At$t=0$, $x=0$and his position at $t=8\min$is,

$$\begin{gathered} x=\mathrm{v}∆{\mathrm{t}}^{\text{'}} \\ =2.2\frac{\mathrm{m}}{\mathrm{s}}×180\mathrm{s} \\ =396\mathrm{m} \end{gathered}$$

Substitute the given values in equation (i).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{avg}}=\frac{∆\mathrm{x}}{∆\mathrm{t}} \\ =\frac{396-0}{360} \\ =1.10\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the average velocity in the time interval $2.00\min$to $8.00\min$is $1.10\mathrm{m}/\mathrm{s}$.

The man is at $t=2min$rest at and has velocity$2.2\mathrm{m}/\mathrm{s}\mathrm{at}\mathrm{t}=8\min$.

Substitute the values in equation (ii) to find the acceleration.

$$\begin{gathered} {\mathrm{a}}_{\mathrm{avg}}=\frac{∆\mathrm{v}}{∆\mathrm{t}} \\ =\frac{2.2-0}{360\mathrm{s}} \\ =0.00611\mathrm{m}/{\mathrm{s}}^2 \\ =6.11\mathrm{mm}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the average acceleration in the time interval $2.00\min$to $8.00\min$is$6.11\mathrm{mm}/{\mathrm{s}}^2$ .

Now entire time interval is$∆t=360s$

Sub time interval in which man is moving,

$$\begin{gathered} ∆{\mathrm{t}}^{\text{'}}=9-5 \\ =4\min \\ =4×60\mathrm{s} \\ =240\mathrm{s} \end{gathered}$$

At $\mathrm{t}=3\min \mathrm{x}=0$ and at $t=9\min$,

$$\begin{gathered} x=v∆t^{\text{'}} \\ =2.2×240 \\ =528\mathrm{m} \end{gathered}$$

Substitute the values in equation (i) to calculate average velocity.

$$\begin{gathered} v_{\mathrm{avg}}=\frac{528-0}{360} \\ =1.47\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the average velocity in the time interval $3.00\min$to $9.00\min$is$1.47m/s$ .

The man is at rest at $t=3\min$and has velocity $v=2.2\mathrm{m}/\mathrm{s}$at $t=9\min$.

So, the average acceleration is same as in part (b) that is $6.11\mathrm{mm}/{\mathrm{s}}^2$.

The following graph of $x$vs $t$, $v$vs $t$indicates how answers to (a) through (d) can be obtained.

Horizontal line near bottom of graph represents man standing at$x=0$at $0<t<300s$and linearly rising line for $300\mathrm{s}<\mathrm{t}<600\mathrm{s}$represents constant velocity motion.

The lines represent answer to part (a) and (c). Slope of these lines would be average velocity for given time intervals.

Man’s average velocity is $1.10\mathrm{m}/\mathrm{s}$. His average acceleration between $2.00\min$to $8.00\min$is . Man’s average velocity and average acceleration time between intervals $3.00\min$to $9.00\min$is $1.47\mathrm{m}/\mathrm{s}$and $6.11\mathrm{mm}/{\mathrm{s}}^2$respectively.

Above graph of $x$vs $t$, $v$vs $t$indicates how answers to (a) through (d) can be obtained.