Post by: luken93 on July 21, 2010, 09:42:58 pm
Determine the acceleration...
Post by: pooshwaltzer on July 21, 2010, 09:47:33 pm
velocity @ t=10 ... 2.85 m/s
UNIFORM acceleration over 10sec = 0.285 m/s^2
Post by: Whatlol on July 21, 2010, 09:50:11 pm
this problem is just a simple plug in your values and execute the calculation (=
Post by: luken93 on July 21, 2010, 09:58:23 pm
haha
Post by: Lighties on July 21, 2010, 10:10:37 pm
t=10, x=10u+50a; u=0, x=50a
Since it's the tenth second, then the change in distance is 50a-40.5a = 8.5a.
28.5 = 8.5a
a=3.35
Sorry if it's a bit hard to see, I'm too lazy to do latex. Uniform acceleration just means that with each passing second the velocity increases, and thus the distance travelled in each second also increases.
Post by: pooshwaltzer on July 21, 2010, 10:13:20 pm
Post by: luken93 on July 21, 2010, 10:27:14 pm
Since it's the tenth second, then the change in distance is 50a-40.5a = 8.5a.I think you mean 50a-40.5a = 9.5a
28.5 = 8.5a
a=3.35
28.5 = 9.5a
a = 3
But thanks alot!
Post by: Lighties on July 23, 2010, 08:31:51 pm
I'm a moron. Ignore my response above.
Since it's the tenth second, then the change in distance is 50a-40.5a = 8.5a.I think you mean 50a-40.5a = 9.5a
28.5 = 8.5a
a=3.35
28.5 = 9.5a
a = 3
But thanks alot!
Haha, we all make mistakes >.>
You're welcome, I did mean 9.5, don't know where the 8 came from. @.@
Post by: pooshwaltzer on July 23, 2010, 08:49:30 pm
A freight train weighing 500 metric tonnes travelling at 150km/h collides into the rear end of a passenger train weighing 200 metric tonnes travelling at 80km/h. Assuming no derailment, what is the net velocity of the combined object post collision? Deceleration of freight train? Acceleration of passenger train?
Post by: Cthulhu on July 23, 2010, 10:11:28 pm
Solving for
After some plugging and chugging:
To do the next bit the best we can hope for is change in velocity