Chapter 2, Problem 24P

To determine

The average acceleration and time of travel for the sprinter

Answer to Problem 24P

  $4.4\frac{\text{m}}{{\text{s}}^{\text{2}}}$ and $2.6\text{s}$

Explanation of Solution

Given:

Initial velocity of the sprinter $=v_i=0\frac{\text{m}}{\text{s}}$

Final velocity of the sprinter $=v_f=11.5\frac{\text{m}}{\text{s}}$

Average acceleration of the sprinter $=a$

Displacement of the sprinter $=\Delta x=15\text{m}$

Time taken $=t$

Formula Used:

Acceleration is given as

  $a=\frac{v^2{}_f-v^2{}_i}{2\Delta x}$

Acceleration is also given as

Acceleration is given as

  $a=\frac{v_f-v_i}{t}$

Where,

  $v_i=$ Initial velocity

  $v_f=$ Final velocity

  $a=$ Acceleration

  $t=$ Time taken

Calculation:

Consider the motion of the sprinter while speeding up.

Average acceleration of the sprinter is given as,

  $a=\frac{v^2{}_f-v^2{}_i}{2\Delta x}$

Substituting the values

  $\begin{array}{l}a=\frac{{11.5}^2-0^2}{2(15)}\\ a=4.4\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Acceleration of the sprinter is also given as,

  $a=\frac{v_f-v_i}{t}$

Substituting the values

  $\begin{array}{l}4.4=\frac{11.5-0}{t}\\ t=\frac{11.5}{4.4}\\ t=2.6\text{s}\end{array}$

Conclusion:

Therefore, average acceleration and time of travel of the sprinter are $4.4\frac{\text{m}}{{\text{s}}^{\text{2}}}$ and $2.6\text{s}$ .