A space vehicle accelerates uniformly from 70 at t = 0 to 172 at t = 10.0. How far did it move between = 2.0 and = 6.0 ?
So I used the forumla d=vt+.5a(t)^2
but for a i got 10.2 since a=(172-70)/10
Now im stuck for v, what do i put for v?
without units, I have no idea what you mean.
To find the distance the space vehicle moved between t = 2.0 s and t = 6.0 s, you can use the formula d = vt + 0.5at^2, as you mentioned.
First, let's calculate the acceleration (a). You correctly found that a = (172 - 70) / 10 = 10.2 m/s^2.
Now, to find the initial velocity (v) of the space vehicle at t = 2.0 s, we need to determine the velocity at t = 0 s (v₀) and the change in time (Δt). Since the vehicle is said to have accelerated uniformly from 70 m/s at t = 0 s, we can assume that it started from rest (v₀ = 0).
Next, we calculate Δt as the difference between the final time (t = 6.0 s) and the initial time (t = 2.0 s): Δt = t₂ - t₁ = 6.0 s - 2.0 s = 4.0 s.
Now we can substitute the values into the formula:
d = vt + 0.5at^2
Since v₀ = 0, the formula simplifies to:
d = 0t + 0.5at^2
Plugging in the values:
d = 0 + 0.5(10.2 m/s^2)(4.0 s)^2
Solving the equation:
d = 0 + 0.5(10.2 m/s^2)(16.0 s^2)
d = 0 + 0.5(163.2 m)
d = 0 + 81.6 m
Therefore, the space vehicle moved a distance of 81.6 meters between t = 2.0 s and t = 6.0 s.