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:
[tex]v=u+at[/tex]
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:
[tex]\begin{gathered} v=0+5(8) \\ \\ v=40\text{ m/s} \end{gathered}[/tex]
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:
[tex]d=v_it+0.5at^2[/tex]
Thus, we have:
[tex]\begin{gathered} d_1=0(8)+0.5(5)8^2 \\ \\ d_1=160\text{ m} \end{gathered}[/tex]
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:
[tex]d=v_it+0.5at^2[/tex]
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:
[tex]\begin{gathered} d_2=40(1800)+0.5(0)1800^2 \\ \\ d_2=72000\text{ m} \end{gathered}[/tex]
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.
