Question

A runner starts from rest and in3 sreaches a speed of . Assume that her speed changed at a constant rate (constant net force).

(a) What was her average speed during this 3 s interval?

(b) How far did she go in this 3 s interval?

Short answer

1. Average speed for the 3 s, $v_{avg}=4\text{\hspace{0.33em}m/s}$
2. Distance travel$s=12\text{\hspace{0.33em}m}$ .

Step-by-step solution

Identification of given data

• Initial velocity $v_i=0$.
• Final velocity $v_f=8\text{\hspace{0.17em}m}/\text{s}$.
• Time $t=3\text{\hspace{0.33em}s}$.

Average speed

Average speed is the ratio of distance traveled per unit of time.

Calculation of the average speed

(a) The average speed of an object is calculated by

$a=\frac{v_f-v_i}{t}$

Here, a = Acceleration , and t = Total time

$\begin{array}{l}a=\frac{8}{3}\\ a=2.66\text{\hspace{0.33em}m}/\text{s}\end{array}$

$\begin{array}{c}s=ut+\frac{1}{2}a{t}^2\\ s=0\times 3\text{\hspace{0.33em}s}+\frac{1}{2}\times 2.66\text{\hspace{0.33em}m}/\text{s}\times {\left(3\text{\hspace{0.33em}s}\right)}^2\\ s=12\text{\hspace{0.33em}m}\end{array}$

role="math" localid="1657104664252" $\begin{array}{c}v=\frac{s}{t}\\ v=\frac{12\text{\hspace{0.33em}m}}{3\text{\hspace{0.33em}s}}\\ =4\text{\hspace{0.33em}m/s}\end{array}$

The average speed will be $\begin{array}{c}4\text{\hspace{0.33em}m/s}\end{array}$.

Calculation for the distance travel

(b) See part a

The distance is traveled $s=12\text{\hspace{0.33em}m}$.