Chapter 3.2, Problem 29PP

To determine

To Calculate: The total distance run by man.

Answer to Problem 29PP

  $4.253\text{km}$

Explanation of Solution

Given:

Velocity from A to B, $v_{AB}=4.5\text{m/s}$

Time elapsed to run from A to B, $t_{AB}=15.0\min$

Acceleration of man up the hill, $a=-0.050{\text{m/s}}^{\text{2}}$ (deceleration)

Time elapsed from B to C, $t_{BC}=90.0\text{s}$

Final velocity at C, $v_C=0$

Formula Used:

Second equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$

Where, s is the displacement, u is the initial velocity, t is the time and a is the acceleration.

For constant velocity, distance can be obtained by:

  $d=vt$

Where, v is the constant velocity and t is the time elapsed.

Calculations:

Find the distance covered from A to B.

  $\begin{array}{c}{d}_{AB}=v_{AB}{t}_{AB}\\ =\left(4.5\text{m/s}\right)\left(15\min \times 60\frac{\text{s}}{\min }\right)\\ =4050\text{m}\end{array}$

Use the second equation of motion to find the distance covered from B to C.

  $\begin{array}{c}{d}_{BC}=\left(4.5\right)\left(90.0\right)+\frac{1}{2}\left(-0.050\right){\left(90.0\right)}^2\\ =405\text{m}-202.5\text{m}\\ =202.5\text{m}\end{array}$

Total distance covered is:

  $\begin{array}{c}{d}_{AB}+d_{BC}=4050\text{m}+202.5\text{m}\\ =4252.5\text{m}\\ =\frac{4252.5\text{m}}{1000\text{m/km}}\\ =4.253\text{km}\end{array}$

Conclusion:

Thus, the total distance covered by man is $4.253\text{km}$ .