Why do I keep getting the wrong answers? A electron moving along the x axis has a position given by x = 10 te-1.3 t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?
I think I may have figured it out...is it 2.8298...m? @iambatman
@JFraser Would you mind helping me out?
\[x = 10te^{-1.3t}\] this correct?
sorry i wasnt paying attention, but I ended up figuring it out :) the answer was the 2.8... Thanks for coming to help though, I really appreciate it!! If you don't mind, would you be willing to help me with a different problem?
Yes I can second that
And sure, if I can figure it out
ok thanks :)
As two trains move along a track, their conductors suddenly notice that they are headed toward each other. Figure 2-27 gives their velocities v as functions of time t as the conductors slow the trains. The figure's vertical scaling is set by vs = 43.0 m/s. The slowing processes begin when the trains are 203 m apart. What is their separation when both trains have stopped?
First ting to do is to find the deceleration of each of the trains using the slope of each line. Acceleration is defined as a change in velocity over a change in time, which is precisely the definition of slope. It gives you Vs = 43.0 m/s, so for the top line for example: a = (0 - 43) / (5 - 0) = -8.6 m/s^2 You can do the same for the other train. Then you can use kinematics equations to solve for the positions. I will again show the top train's line. Vf^2 = Vi^2 + 2a(Xf - Xi) Vf = 0 m/s Vi = Vs = 43.0 m/s a = -8.6 m/s^2 Xi = 0 m Xf = ? what we need 0 = 43^2 + (2)(-8.6)(Xf - 0) -1849 = -17.2 Xf Xf = 107.5 m You can do the same for the other train. Afterwards, just find the difference between the positions based on the initial separation. If we let the top line train start at 0 m, then it finishes slowing at 107.5 m. When you find the bottom line train distance, subtract it from 203 m since it is starting at the other end. The two values you have (107.5 and the other distance) will have a difference which is equal to the separation of the trains.
This makes a lot of sense :) thank you
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