Chapter 2, Problem 17P

To determine

The acceleration in (a) ${\text{m/s}}^{\text{2}}$ (b) ${\text{km/h}}^{\text{2}}$ .

Answer to Problem 17P

Solution:

In $m/s^2=6.24m/s^2$

In $km/h^2=85263km/h^2$

Explanation of Solution

Given:

A sprinter accelerates from rest to 9.00m/s in 1.38 s.

  $\begin{array}{l}u=0\\ v=9\text{m/s}\\ t=1.38\text{s}\end{array}$

Formula Used:

  $\text{Acceleration}=\frac{\text{change in velocity}}{\text{change in time}\text{.} }$

Calculation:

Convert velocity from m/s to km/h.

  $\begin{array}{l}v=9.00\text{ m/s}\\ v=\frac{9\times 3600}{1000}=32.4\text{ km/h}\end{array}$

The time taken was 1.38 s.

Convert it into hours

  $=\frac{\text{1}\text{.38}}{\text{3600}}\text{h = 0}\text{.00038 h}$

a) Acceleration in ${\text{m/s}}^{\text{2}}$

  $\text{Acceleration}= \frac{\left(\text{final velocity}-\text{initial velocity}\right)}{t}$

  $\begin{array}{l}\text{Average acceleration}=\frac{9-0}{1.38}\\ =6.52{\text{ m/s}}^{\text{2}}\end{array}$

b) Acceleration in ${\text{km/h}}^{\text{2}}$

  $\text{Acceleration}= \frac{\left(\text{final velocity}-\text{initial velocity}\right)}{t}$

  $\begin{array}{l}\text{Average acceleration}=\frac{32.4-0}{0.00038}\\ =85263km/h^2\end{array}$