Question

An object has an initial forward velocity of 25 m/s and a constant acceleration of -1.5 mis.

How far does it travel in 5.0 seconds?

Answer 1

The object with initial velocity of 25 m/s and a constant acceleration of -1.5 m/s² travels 106.25 meters in 5.0 seconds.

Step-by-step explanation:

To calculate the distance an object travels under constant acceleration, we can use the kinematic equation for displacement, which is:

s = ut + ½at²

Where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Given that the initial velocity (u) is 25 m/s, the acceleration (a) is -1.5 m/s², and the time (t) is 5.0 seconds, we can plug these values into the equation:

s = (25 m/s)(5.0 s) + ½(-1.5 m/s²)(5.0 s)²

s = 125 m - ½(1.5 m/s²)(25 s²)

s = 125 m - 18.75 m

s = 106.25 meters

Therefore, the object travels 106.25 meters in 5.0 seconds.

Answer 2

Answer: 106. 25 m

Step-by-step explanation:

In this situation the following equations will be useful:

`V=V_(o)+at` (1)

`V^(2)=V_(o)^(2)+2ad` (2)

Where:

`V` is the object’s final velocity

`V_(o)=25 m/s` is the object’s initial velocity

`a=-1.5 m/s^(2)` is the object's acceleration

`t=5 s` is the time

`d` is the distance traveled

Finding
`V` from (1):

`V=25 m/s+(-1.5 m/s^(2))(5 s)` (3)

`V=17.5 m/s` (4)

Substituting (4) in (2):

`(17.5 m/s)^(2)=(25 m/s)^(2)+2(-1.5 m/s^(2))d` (5)

Finding
`d`:

`d=106.25 m`