If a shopper applies a 9.0 N force for 8.00 seconds to stop a loaded 41.5 kg cart from rolling down a slope, how fast was the cart initially rolling away?

Use the equation:
[tex] F \cdot t = m \cdot \Delta v [/tex]

Possible answers:
A. [tex]1.7 \, \text{m/s}[/tex]
B. [tex]36.9 \, \text{m/s}[/tex]
C. [tex]58.5 \, \text{m/s}[/tex]
D. [tex]0.58 \, \text{m/s}[/tex]

To find out how fast the cart was rolling away, we can use the impulse-momentum formula. According to this formula:

[tex]\[ F \times t = m \times \Delta v \][/tex]

Where:
- [tex]$ F $[/tex] is the force applied (in Newtons)
- [tex]$ t $[/tex] is the time duration over which the force is applied (in seconds)
- [tex]$ m $[/tex] is the mass of the cart (in kilograms)
- [tex]$ \Delta v $[/tex] is the change in velocity (in meters per second)

Given values are:
- Force ([tex]$ F $[/tex]): 9.0 N
- Time ([tex]$ t $[/tex]): 8.00 seconds
- Mass ([tex]$ m $[/tex]): 41.5 kg

We need to solve for the change in velocity ([tex]$ \Delta v $[/tex]):

1. Calculate the change in velocity:

[tex]\[
\Delta v = \frac{F \times t}{m}
\][/tex]

2. Substitute the given values:

[tex]\[
\Delta v = \frac{9.0 \, \text{N} \times 8.00 \, \text{s}}{41.5 \, \text{kg}}
\][/tex]

3. Perform the multiplication in the numerator:

[tex]\[
9.0 \, \text{N} \times 8.00 \, \text{s} = 72.0 \, \text{Ns}
\][/tex]

4. Divide by the mass:

[tex]\[
\Delta v = \frac{72.0 \, \text{Ns}}{41.5 \, \text{kg}}
\][/tex]

5. Calculate the result:

[tex]\[
\Delta v \approx 1.73 \, \text{m/s}
\][/tex]

The closest listed option to this calculated change in velocity is [tex]$ 1.7 \, \text{m/s} $[/tex]. Thus, the cart was initially rolling away approximately at a speed of [tex]$ 1.7 \, \text{m/s} $[/tex].