Question

A driver speeds up from 1.9 m/s to 4.4 m/s in 9.6 s, with a constantacceleration. The traveled distance in meters is:

Answer 1

Recall the following formula for motion at constant acceleration:

$2ad=v^2_f-v^2_i$

Where a represents the acceleration, and d represents the distance traveled while accelerating from the initial velocity v_i to the final velocity v_f.

First, solve for the distance d:

$d=\frac{v^2_f-v^2_i_{}}{2a}$

The values of v_f and v_i are known, and the value of a can be calculated using the definition of acceleration:

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

Substitute v_f=4.4 m/s, v_i=1.9 m/s and t=9.6s to find the acceleration:

$\begin{gathered} a=\frac{4.4\frac{m}{s}-1.9\frac{m}{s}}{9.6s} \\ =\frac{2.5\frac{m}{s}}{9.6s} \\ =0.2604\frac{m}{s^2} \end{gathered}$

Next, substitute a=0.2604 m/s^2 as well as v_f=4.4 m/s and v_i=1.9 m/s into the formula for d to find the traveled distance:

$\begin{gathered} d=\frac{(4.4\frac{m}{s})^2-(1.9\frac{m}{s})^2}{2(0.2604)\frac{m}{s^2}} \\ =\frac{19.36\frac{m^2}{s^2}-3.61\frac{m^2}{s^2}}{0.52\frac{m}{s^2}} \\ =\frac{15.75\frac{m^2}{s^2}}{0.5208\frac{m}{s^2}} \\ =30.2m \end{gathered}$

Therefore, the traveled distance in meters is equal to:

$30.2$