A space vehicle accelerates uniformly from 50m/s at t=0 to 182m/s at t=1 0.0s.
How far did it move between t=2.0s and t=6.0s?
a = change in velocity/change in time
= (182-50)/10
d = Vi t + .5 a t^2
doesn't help wtf
To find the distance the space vehicle moved between t = 2.0s and t = 6.0s, we need to first determine the acceleration of the vehicle. The given information tells us the initial velocity (u) and final velocity (v) at t = 0s and t = 10.0s, respectively.
Using the equation for uniformly accelerated motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Rearranging the equation, we can solve for the acceleration (a):
a = (v - u) / t
Substituting the values:
a = (182 m/s - 50 m/s) / (10.0 s - 0 s)
a = 13.2 m/s²
Now that we know the acceleration, we can find the distance traveled between t = 2.0s and t = 6.0s using the equation for distance in uniformly accelerated motion:
s = ut + 0.5at²
Since we are given only the initial velocity and not the initial time, we need to find the distance traveled between t = 2.0s and t = 6.0s as if the initial velocity were 0 m/s. Then we can subtract the distance traveled at t = 2.0s to find the final answer.
Finding the distance traveled from t = 0s to t = 6.0s:
s1 = 0.5at²
s1 = 0.5 * 13.2 m/s² * (6.0 s)²
s1 = 237.6 m
Finding the distance traveled from t = 0s to t = 2.0s:
s2 = 0.5at²
s2 = 0.5 * 13.2 m/s² * (2.0 s)²
s2 = 13.2 m
Finally, we can find the distance traveled between t = 2.0s and t = 6.0s by subtracting s2 from s1:
Distance traveled = s1 - s2
Distance traveled = 237.6 m - 13.2 m
Distance traveled = 224.4 m
Therefore, the space vehicle moved 224.4 meters between t = 2.0s and t = 6.0s.