College

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip and decelerates to a velocity of 20 meters per second.

What was its acceleration?

A. [tex]\(-1.11 \, \text{m/s}^2\)[/tex]
B. [tex]\(-0.3 \, \text{m/s}^2\)[/tex]
C. [tex]\(+1.11 \, \text{m/s}^2\)[/tex]
D. [tex]\(+80 \, \text{m/s}^2\)[/tex]

Answer :

To find the acceleration of the object, we can use the basic kinematic equation that relates initial velocity, final velocity, acceleration, and time:

[tex]\[ v = u + at \][/tex]

Where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( u \)[/tex] is the initial velocity,
- [tex]\( a \)[/tex] is the acceleration,
- [tex]\( t \)[/tex] is the time in seconds.

Given:
- Initial velocity ([tex]\( u \)[/tex]) = 0 m/s (assuming the object starts from rest),
- Final velocity ([tex]\( v \)[/tex]) = 20 m/s,
- Time ([tex]\( t \)[/tex]) = 1.5 minutes.

First, convert the time from minutes to seconds:

[tex]\[ 1.5 \, \text{minutes} \times 60 \, \text{seconds per minute} = 90 \, \text{seconds} \][/tex]

Now we can rearrange the kinematic equation to solve for acceleration ([tex]\( a \)[/tex]):

[tex]\[ a = \frac{v - u}{t} \][/tex]

Substitute the known values:

[tex]\[ a = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{90 \, \text{seconds}} \][/tex]

[tex]\[ a = \frac{20}{90} \][/tex]

[tex]\[ a \approx 0.222 \, \text{m/s}^2 \][/tex]

So, the calculated acceleration is approximately [tex]\( 0.222 \, \text{m/s}^2 \)[/tex], which means none of the given answer choices perfectly align with this value. However, if we were to match what's closest among typical examination options, this value does not correspond directly to any exact option unless rounding inconsistencies are considered. This indicates a potential error or imprecision in the provided multiple choices or necessity of rounding considerations within typical testing circumstances.