A space vehicle accelerates uniformly from 75m/s at t = 0 to 170m/s at t = 10.0 s. How far did it move between t = 2.0 s and t = 6.0 s ?

Question

anonymous · Answer

can u help me im only in 6TH grade

anonymous · Answer

and i do not know

anonymous · Answer

Your initial velocity is 75m/s at t=0. Your final velocity is 170m/s at t=10.

170-75 = 95 m/s, which is the change in velocities from t=0 to t=10.

As acceleration is basically change in velocity/time, we divide 95m/s by 10s, yielding a constant acceleration value of 9.5m/s^2.

Using this, we calculate the value of the velocity of the spacecraft at t=2 and t=6 respectively :

9.5m/s^2 * 2 + 75 (initial velocity) = 94m/s at t=2.
9.5m/s^2 * 6 + 75 = 132m/s at t=6.

From here on, it gets easier. You know that the distance covered is essentially the area UNDER a velocity time graph, right? So, on the vertical axis of the velocity "triangle", we have 132, and 94 m/s. On the horizontal x axis, we have t=2 and t=6.

Therefore the "width" of our triangle is 4s, the change in time (6-2). The height would be the change in velocity from t=2 to t=6, which is 38. (132-94)

From there on, it's a simple matter of using the formula of an area of a triangle, h*b*.5

So, 38 * 4 * 0.5 which equals 76m.

Answer => 76m.

I hope this helped you. I apologize for overcomplicating the process, but feel free to contact me in case you still don't understand.

anonymous · Answer

So that's not the answer? Oooh. Right.

anonymous · Answer

Do you have any other ideas?

anonymous · Answer

I'm sorry, I really need to sleep. I'll answer that question tomorrow. I hope you don't mind!