Could someone help me with this question
An object moves in one dimensional motion with constant acceleration a = 3.2 m/s2. At time t = 0 s, the object is at x0 = 4.3 m and has an initial velocity of v0 = 3.4 m/s.
How far will the object move before it achieves a velocity of v = 8.2 m/s?
@NellyXO For constant acceleration, we have v(t) = v_0 + at. Use this to solve for the time (t) for which velocity = 8.2 m/s.
Then its position at time t is x(t) = (1/2)at^2 + v0*t + x0. Plug in the time t that you obtained earlier to get the position.
@MITer94 following these equations I got 4.7m. Is that correct?
@NellyXO I got a different value. What value of t did you get (t = the time until velocity = 8.2 m/s)?
@MITer94 I got t=1.5
@NellyXO that’s what I got. Then
x (1.5) = (1/2)3.2(1.5)^2 + 3.4*1.5 + 4.3 = 13 m.
But we are only concerned on how far the object moved. Its displacement is 13 m - 4.3 m = 8.7 m. Because it only moves in the +x direction, 8.7 m is also the distance it moves.
@MITer94 thank you so much! You’ve been very helpful.
Just a heads up, you’re not supposed to post homework help questions on CC.
Closing thread for the reason above.