Question

Hello, Here is a physics question I am having difficulty with, thanks.

Answer 1

Given:

Acceleration for the first 8 seconds = 5 m/s²

Time traveled after 8 seconds at a constant rate = 30 minutes

Let's find how far they traveled before dropping off.

Apply the motion formula to find the velocity traveled in the first 8 seconds:

`v=u+at`

Where:

u is the initial velocity = 0 m/s

a is the acceleration = 5 m/s²

t is the time in seconds = 8 seconds

We have:

`\begin{gathered} v=0+5(8) \\ \\ v=40\text{ m/s} \end{gathered}`

The velocity after the first 8 seconds is 40 m/s.

Now, apply the formula to find the distance traveled in the first 8 seconds:

`d=v_it+0.5at^2`

Thus, we have:

`\begin{gathered} d_1=0(8)+0.5(5)8^2 \\ \\ d_1=160\text{ m} \end{gathered}`

Therefore, for the first 8 seconds, the covered distance was 160 meters.

Now, since the final velocity after 8 seconds is 40 m/s, the initial velocity, when it starts traveling at a constant rate, will be 40 m/s.

Apply the formula to find the distance covered for the next 30 minutes:

`d=v_it+0.5at^2`

Where:

t is the time in seconds = 30 x 60 = 1800 seconds

a is the acceleration

vi is the initial velocity = 40 m/s

Since it traveled at a constant rate, there will be no acceleration.

a = 0

Hence, we have:

`\begin{gathered} d_2=40(1800)+0.5(0)1800^2 \\ \\ d_2=72000\text{ m} \end{gathered}`

The distance traveled after 30 minutes is 72000 meters.

Therefore, the total distance will be:

d1 + d2 = 160 + 72000 = 72160 meters

Therefore, they traveled for 72160 meters before dropping off.

ANSWER:

72160 meters.