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=(4.4(m)/(s)-1.9(m)/(s))/(9.6s) \\ =(2.5(m)/(s))/(9.6s) \\ =0.2604(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=((4.4(m)/(s))^2-(1.9(m)/(s))^2)/(2(0.2604)(m)/(s^2)) \\ =(19.36(m^2)/(s^2)-3.61(m^2)/(s^2))/(0.52(m)/(s^2)) \\ =(15.75(m^2)/(s^2))/(0.5208(m)/(s^2)) \\ =30.2m \end{gathered}`

Therefore, the traveled distance in meters is equal to:

`30.2`