Given:
Velocity at t = 0: 80 m/s
Velocity at t = 10.0 s: 167 m/s
Let's find the distance it covered between t = 2.0s and t = 6.0 s
Apply the kinematics formula:
[tex]v=u+at[/tex]
Where:
v is the final velocity ==> 167 m/s
u is the initial velocity ==> 80 m/s
a is the acceleration
t is the time ==> 10.0 - 0 = 10.0 s
Now, let's find the acceleration.
Rewrite the formula for a:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ a=\frac{167-80}{10} \\ \\ a=\frac{87}{10} \\ \\ a=8.7m/s^2 \end{gathered}[/tex]
The acceleration is 8.7 m/s²
To find the distance, apply the formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]
Thus, we have:
• At t = 2.0 s:
[tex]\begin{gathered} d_1=80(2)+\frac{1}{2}(8.7)(2)^2 \\ \\ d_1=160+17.4 \\ \\ d_1=177.4\text{ m} \end{gathered}[/tex]
• At t = 6.0 s:
[tex]\begin{gathered} d_2=ut+\frac{1}{2}at^2 \\ \\ d_2=80(6)+\frac{1}{2}(8.7)(6)^2 \\ \\ d_2=480+156.6 \\ \\ d_2=636.6\text{ m} \end{gathered}[/tex]
To find the distance traveled between t = 2.0 s and t = 6.0 s, we have:
d = d2 - d1 = 636.6 m - 177.4 m
d = 459.2 m
Therefore, the distance traveled betwen t = 2.0 s and t = 6.0 s is 459.2 m
ANSWER:
459.2 m
