Question

A runner achieves a velocity of 11.1 m/s 9 seconds after he begins. What is his acceleration? What distance did he cover?

Answer 1

$Acceleration=\frac { Final\quad Velocity\quad -\quad Initial\quad Velocity }{ Time }$

Therefore:

$Acceleration=\frac { 11.1\quad m/s\quad -\quad 0\quad m/s }{ 9\quad seconds } \\ \\ =\frac { 11.1\quad m/s }{ 9\quad seconds } \\ \\ =1.23\quad m/{ s }^{ 2 }\quad (2\quad decimal\quad places)$

As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds  = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.

Answer 2

Explanation:

Acceleration=

Time

FinalVelocity−InitialVelocity

Therefore:

\begin{gathered}Acceleration=\frac { 11.1\quad m/s\quad -\quad 0\quad m/s }{ 9\quad seconds } \\ \\ =\frac { 11.1\quad m/s }{ 9\quad seconds } \\ \\ =1.23\quad m/{ s }^{ 2 }\quad (2\quad decimal\quad places)\end{gathered}

Acceleration=

9seconds

11.1m/s−0m/s

=

9seconds

11.1m/s

=1.23m/s

2

(2decimalplaces)

As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.