Question

An emu moving with constant acceleration covers the distance between two points that are 92 m

apart in 6.5s. Its speed as it passes the second point is 14 m/s.

Answer 1

a)
`V_(o)=14.30 m/s`

b)
`a=-0.046 m/s^(2)`

Step-by-step explanation:

The complete question is written below:

An emu moving with constant acceleration covers the distance between two points that are 92 m apart in 6.5s. Its speed as it passes the second point is 14 m/s. What are (a) its speed at the first point and (b) its acceleration?

Since we are talking about constant acceleration, we can use the following equations:

`d=x-x_(o)=((V_(o)-V)/(2))t` (1)

`V=V_(o)+at` (2)

Where:

`d=92 m` is the distance between the two points

`V_(o)` is the velocity of the emu at the first point

`V=14 m/s` is the velocity of the emu at the second point

`t=6.5 s` is the time it takes to the emu to cover the distance
`d`

`a` is the emu's constant acceleration

Knowing this, let's begin with the answers:

a) Speed at the first point

In this situation wi will use equation (1):

`d=((V_(o)-V)/(2))t` (1)

Finding
`V_(o)`:

`V_(o)=(2d)/(t)-V` (3)

`V_(o)=(2(92 m))/(6.5 s)-14 m/s` (4)

`V_(o)=14.30 m/s` (5)

b) Emu's acceleration

Now we will substitute (5) in equation (2):

`14 m/s=14.30 m/s+a(6.5 s)` (6)

Finding
`a`:

`a=-0.046 m/s^(2)` (7) This means the emu is decreasing its speed at a constant rate.