High School

A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling at 120 meters per second 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 a basic kinematic equation that relates acceleration, initial velocity, final velocity, and time. The formula we need is:

[tex]\[ \text{acceleration} = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} \][/tex]

Given:
- Initial velocity ([tex]\(v_i\)[/tex]) = 120 meters per second
- Final velocity ([tex]\(v_f\)[/tex]) = 20 meters per second
- Time = 1.5 minutes

First, we need to convert the time into seconds since velocities are in meters per second. There are 60 seconds in a minute, so:

[tex]\[ \text{Time} = 1.5 \times 60 = 90 \text{ seconds} \][/tex]

Now, plug these values into the formula:

[tex]\[ \text{acceleration} = \frac{20 \, \text{m/s} - 120 \, \text{m/s}}{90 \, \text{s}} \][/tex]

[tex]\[ \text{acceleration} = \frac{-100 \, \text{m/s}}{90 \, \text{s}} \][/tex]

[tex]\[ \text{acceleration} \approx -1.11 \, \text{m/s}^2 \][/tex]

So, the acceleration of the object is approximately [tex]\(-1.11 \, \text{m/s}^2\)[/tex].

Therefore, the correct answer from the given options is:

[tex]\(-1.11 \, \text{m/s}^2\)[/tex]