Answer:
[tex]\huge\boxed{\sf S = 18.75 \ meters}[/tex]
Explanation:
Given Data:
Initial Velocity = Vi = 0 m/s (rest)
Final Velocity for 6 seconds = Vf = 25 m/s
Time (1) = T1 = 6 seconds
Time (2) = T2 = 3 seconds
Required:
Distance for 3 seconds = S = ?
Solution:
For 6 seconds, the acceleration will be:
[tex]\displaystyle a = \frac{Vf-Vi}{t} \\\\a = \frac{25 - 0}{6} \\\\a = 25 / 6\\\\\boxed{a = 4.167 \ m/s^2}[/tex]
Since, acceleration is constant, it will be the same at 3 seconds as well.
Using second equation of motion to find Distance (S) with time being 3 seconds:
[tex]\displaystyle S= Vit+\frac{1}{2} at^2\\\\S = (0)(3)+ \frac{1}{2} (4.167)(3)^2\\\\S = \frac{1}{2} (4.167)(9)\\\\S = \frac{37.5}{2} \\\\\boxed{S = 18.75 \ meters}\\\\\rule[225]{225}{2}[/tex]
Hope this helped!
