Question

From t=0 to$\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{min}$, a man stands still, and from$\mathbf{t}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{min}$to$\mathbf{t}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{min}$, he walks briskly in a straight line at a constant speed of$\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{m}\mathbf{/}\mathbf{s}$. What are a) his average velocity${\mathbf{V}}_{\mathbf{avg}}$b)his average acceleration ${\mathbf{a}}_{\mathbf{avg}}$in time interval$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{min}$to$\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{min}$? What are c)${\mathbf{V}}_{\mathbf{avg}}$d)${\mathbf{a}}_{\mathbf{avg}}$in time interval$\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{min}$to$\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{min}$? e) Sketch x vs t and v vs t and indicate how the answers of (a) through (d) can be obtained from the graphs.

Short answer

(a) Man’s average velocity in the time interval $2.00\min$to$8.00\min$ is $1.10\mathrm{m}/\mathrm{s}$.

(b) Man’s average acceleration in the time interval role="math" localid="1656400586469" $2.00\min$to role="math" localid="1656400621416" $8.00\min$is$6.11\mathrm{mm}/{\mathrm{s}}^2$.

(c) Man’s average velocity in the time interval role="math" localid="1656400636736" $3.00\min$to role="math" localid="1656400666466" $9.00\min$is $1.47\mathrm{m}/\mathrm{s}$.

(d) Man’s average acceleration in the time interval$3.00\min$ to$9.00\min$isdata-custom-editor="chemistry" $6.11\mathrm{m}/{\mathrm{s}}^2$ .

(e) Graphs of $x$vs$t$ , $v$vs$t$ indicate how answers to (a) through (d) can be obtained.

Step-by-step solution

Given Data

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$2.20\mathrm{m}/\mathrm{s}$ .

Understanding the average velocity and average acceleration

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 $t_2$.

(a) Determination of the average velocity between 2 min to 8 min.

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

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

Sub-interval in which man is moving,

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

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

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

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 $1.10\mathrm{m}/\mathrm{s}$.

(b) Determination of the average acceleration between 2 min to 8 min.

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$6.11\mathrm{mm}/{\mathrm{s}}^2$ .

(c) Determination of the man’s average velocity between 3 min to 9 min.

Now entire time interval is$∆t=360s$

Sub time interval in which man is moving,

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

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

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

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$1.47m/s$ .

(d) Determination of the man’s average acceleration between 3 min to 9 min.

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 $6.11\mathrm{mm}/{\mathrm{s}}^2$.

(e) Sketch x vs t and v vs t

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.