quick Motion Q

[luken93]

A car, which is accelerating uniformly from rest, travels a distance of 28.5m in the tenth second of its motion.
Determine the acceleration...

[pooshwaltzer]

velocity @ t=0 ... 0 m/s
velocity @ t=10 ... 2.85 m/s

UNIFORM acceleration over 10sec = 0.285 m/s^2

[Whatlol]

just remember the equation for acceleration a = change in velocity / change in time

this problem is just a simple plug in your values and execute the calculation (=

[luken93]

except the answer is 3m/s
haha

[Lighties]

t=9, x=9u+40.5a; u=0, x=40.5a

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.

[pooshwaltzer]

I'm a moron. Ignore my response above.

[luken93]

Since it's the tenth second, then the change in distance is 50a-40.5a = 8.5a.
28.5 = 8.5a
a=3.35

I think you mean 50a-40.5a = 9.5a
28.5 = 9.5a
a = 3

But thanks alot!

[Lighties]

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.
28.5 = 8.5a
a=3.35

I think you mean 50a-40.5a = 9.5a
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. @.@

[pooshwaltzer]

Here's a fun question,

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?

[Cthulhu]

Conservation of momentum


Solving for

After some plugging and chugging:



To do the next bit the best we can hope for is change in velocity