The 70.0-kg swimmer in Figure 7.44 starts a race with an initial velocity of 1.25 m/s and exerts an average force of 80.0 N backward with his arms during each 1.80 m long stroke.

(a) What is his initial acceleration if water resistance is 45.0 N?

(b) What is the subsequent average resistance force from the water during the 5.00 s it takes him to reach his top velocity of 2.50 m/s?

(c) Discuss whether water resistance seems to increase linearly with velocity.

The force is a vector quantity which is defined as the mass times acceleration.

Mathematically,

F = ma

Here, F is the force, m is the mass of the swimmer, and a is the acceleration of the swimmer.

Free body diagram of the swimmer

Here, F_net is the resultant force, F is the force exerted by the swimmer, and F_r is the water resistance.

(a)

The resultant force is,

F_net = F - F_r

Here, F is the average force the swimmer exerts (F = 80 N), and F_r is the water resistance (F_r = 45 N).

Putting all known values,

$\begin{array}{rcl}{F}_{net}& =& \left(80.0\text{N}\right)-\left(45.0\text{N}\right)\\ & =& 35\text{N}\\ & & \end{array}$

The resultant force is,

$F_{net}=ma$

Here, m is the mass of the swimmer (m = 70 kg), and$F_{net}$ is the resultant force$F_{net}=35 N$ , and a is the initial acceleration of the swimmer.

The expression for the initial acceleration of the swimmer is,

$a=\frac{F_{net}}{m}$

Putting all known values,

$\begin{array}{rcl}a& =& \frac{35.0\text{N}}{70.0\text{kg}}\\ & =& 0.5\text{m}/{\text{s}}^2\\ & & \end{array}$

Therefore, the initial acceleration of the swimmer is $0.5m/s^2$.

(b)

The acceleration of the swimmer is,

$a=\frac{v-u}{t}$

Here, v is the final velocity of the swimmer$\left(v=2.50\text{m}/\text{s}\right)$ , u is the initial velocity of the swimmer$\left(u=1.25\text{m}/\text{s}\right)$ , and t is the time$\left(t=5.00\text{s}\right)$ .

Putting all known values,

$\begin{array}{rcl}a& =& \frac{\left(2.50\text{m}/\text{s}\right)-\left(1.25\text{m}/\text{s}\right)}{\left(5.00\text{s}\right)}\\ & =& 0.25\text{m}/{\text{s}}^2\\ & & \end{array}$

The net force is,

$F_{net}=ma$

Here,$F_{net}$is the net resultant force, m is the mass of the swimmer (m = 70 kg), and a is the acceleration of the swimmer$\left(a=0.25\text{m}/{\text{s}}^2\right)$

Putting all known values,

$\begin{array}{rcl}{F}_{net}& =& \left(70\text{kg}\right)\times \left(0.25\text{m}/{\text{s}}^2\right)\\ & =& 17.5\text{N}\\ & & \end{array}$

The net resultant force is,

$F_{net}=F-F_r$

Here, F is the average force the swimmer exerts$\left(F=80.0\text{N}\right)$ , and$F_r$ is the water resistance, and$F_{net}$ is the net resultant force$\left(F_{net}=17.5\text{N}\right)$ .

The expression for the water resistance is,

$F_r=F-F_{net}$

Putting all known values,

$\begin{array}{rcl}{F}_r& =& \left(80.0\text{N}\right)-\left(17.5\text{N}\right)\\ & =& 62.5\text{N}\\ & & \end{array}$

Therefore, the required average resistance force from the water during$5.00s$ is$62.5N$ .

(c)

No, the change in water resistance is not linear with respect to time.