Chapter 3.2, Problem 31PP

To determine

To Calculate: The acceleration during the last portion of the training run.

Answer to Problem 31PP

  $a=0.077{\text{m/s}}^2$

Explanation of Solution

Given:

Total distance $d=5\text{km = 5000 m}$

Velocity $=4.3\text{m/s}$

Time for first part $=19\text{min}$

Time for second part $=19.4\text{s}$

Formula used:

Second equation of motion:

  $d_f-d_i=ut+\frac{1}{2}a{t}^2$

Here,

  $d_f$ is the final position

  $d_i$ is the initial position

  $u$ is the initial velocity

  $t$ is the time

  $a$ is the acceleration

Conversion factor:

  $1\min =60\text{s}$

Calculation:

Distance covered the first $19\text{min}$ can be calculated as:

  $\begin{array}{l}{d}_1=v{t}_1\\ d_1=4.3\times 19\times 60\\ d_1=4902\text{m}\end{array}$

Distance left to cover in second part

  $\begin{array}{l}{d}_2=d-d_1\\ d_2=5000-4902\text{m}\\ d_2=98\text{m}\end{array}$

Acceleration for the last part can be calculated as:

  $\begin{array}{l}{d}_2=ut+\frac{1}{2}a{t}^2\\ 98=4.3\times 19.4+0.5\times {\left(19.4\right)}^{2}a\\ a=\frac{98-4.3\times 19.4}{0.5\times {\left(19.4\right)}^2}\\ a=0.077{\text{m/s}}^2\end{array}$

Conclusion:

  $a=0.077{\text{m/s}}^2$ is the required acceleration.