Chapter 3.1, Problem 14SSC

(a)

To determine

To Calculate: The average velocity of canoe.

Answer to Problem 14SSC

  $1.0\text{m/s}$ downstream

Explanation of Solution

Given:

Upstream velocity, $v_u=2.0\text{m/s}$

Downstream velocity, $v_d=-4.0\text{m/s}$ (considering negative for downstream)

Time elapsed, $t=4.\text{0}\text{s}$ for upstream and downstream.

Formula Used:

Average velocity can be obtained:

  $v_{avg}=\frac{s_{net}}{t}$

Here, $s_{net}$ is the net displacement and t is the time.

Calculations:

Displacement of canoe upstream:

  $\begin{array}{c}{s}_u=v_{u}t\\ =\left(2.0\right)\left(4.0\right)\\ =8.0\text{m}\end{array}$

Displacement of canoe downstream:

  $\begin{array}{c}{s}_d=v_{d}t\\ =\left(-4.0\right)\left(4.0\right)\\ =-16.0\text{m}\end{array}$

Net displacement:

  $\begin{array}{c}{s}_{net}=s_u+s_d\\ =\text{8}\text{.0}\text{m}-\text{16}\text{.0}\text{m}\\ \text{=-8}\text{.0}\text{m}\end{array}$

Here, negative sign indicated that the net displacement is downstream.

Average velocity can be calculated as follows:

  $\begin{array}{c}{v}_{avg}=\frac{-8.0\text{m}}{8.0\text{s}}\\ =-1.0\text{m/s}\end{array}$

Conclusion:

Thus, the average velocity of the canoe is $1.0\text{m/s}$ downstream.

(b)

To determine

To Calculate: The average acceleration of canoe.

Answer to Problem 14SSC

  $-0.75{\text{m/s}}^{\text{2}}$

Explanation of Solution

Given:

Upstream velocity, $v_u=2.0\text{m/s}$

Downstream velocity, $v_d=-4.0\text{m/s}$ (considering negative for downstream)

Time elapsed, $t=4.\text{0}\text{s}$ for upstream and downstream.

Formula Used:

First equation of motion:

  $v=u+at$

Where, v is the final velocity, u is the initial velocity, t is the time and a is the acceleration.

Calculations:

Average acceleration can be calculated as follows:

  $\begin{array}{l}a=\frac{v-u}{t}\\ a=\frac{-4.0-2.0}{8.0}{\text{m/s}}^{\text{2}}\\ a=-0.75{\text{m/s}}^{\text{2}}\end{array}$

Negative sign indicates deceleration.

Conclusion:

Thus, the average acceleration of canoe is $-0.75{\text{m/s}}^{\text{2}}$ .