Question

a car, initially traveling 110 ft/s steadily slows to a stop in 5.9 s. determine all unknowns and answer the following question.

$s_i = \boldsymbol{10}$ ft/s
$s = \boldsymbol{5}$ ft/s
$s_f = 0$ ft/s
$d = \boldsymbol{25}$ ft
$t = \boldsymbol{5}$ s

how far did the car travel during this time?

$\boldsymbol{25}$ ft/s

Explanation:

Step1: Identify the correct formula

For an object with constant acceleration (here, deceleration as it's slowing down), the distance traveled \( d \) can be found using the average velocity formula. The average velocity \( v_{avg} \) is the average of the initial velocity \( v_i \) and final velocity \( v_f \). The formula for distance is \( d = v_{avg} \times t \), where \( v_{avg}=\frac{v_i + v_f}{2} \).

Given: \( v_i = 110 \, \text{ft/s} \), \( v_f = 0 \, \text{ft/s} \) (comes to stop), \( t = 5.9 \, \text{s} \).

Step2: Calculate average velocity

First, find the average velocity:
\( v_{avg}=\frac{v_i + v_f}{2}=\frac{110 + 0}{2}= 55 \, \text{ft/s} \)

Step3: Calculate distance traveled

Now, use the distance formula \( d = v_{avg} \times t \). Substitute \( v_{avg} = 55 \, \text{ft/s} \) and \( t = 5.9 \, \text{s} \):
\( d=55\times5.9 = 324.5 \, \text{ft} \)

Answer:

The car traveled \( \boldsymbol{324.5} \) feet.