A toy train is pushed forward and released at xi = 2.5m with a speed of 2.5m/s . It rolls at a steady speed for 3.0s , then one wheel begins to stick. The train comes to a stop 9.4m from the point at which it was released.

What is the train's acceleration after its wheel begins to stick?

Question

A toy train is pushed forward and released at xi = 2.5m with a speed of 2.5m/s . It rolls at a steady speed for 3.0s , then one wheel begins to stick. The train comes to a stop 9.4m from the point at which it was released.

What is the train's acceleration after its wheel begins to stick?

nottim · Answer

I have to go to my job in 20 minutes, but I will be back within 3 hours.

anonymous · Answer

can u help me with my last question please

anonymous · Answer

these questions involving steady forces and thus steady accelerations (a)
can be solved by using 

v^2 - u^2 = 2 a s

where v is final velocity and u is initial velocity
a is acceleration 
s is distance

The other common problem is when you are told the acceleration is constant and you are given the time, so v = u + a t

nottim · Answer

so this isn't a two step problem @douglaswinslowcooper

anonymous · Answer

I am not sure what "two-step" means. Use my v^2 - u^2 equation to solve.

nottim · Answer

only that right? because I dunno the final velocity BEFORE it hit the sticky part, and the initial velocity AFTER it hit the sticky part (but these are the same values)

anonymous · Answer

v = "steady speed" of ....

u = "comes to a stop."

nottim · Answer

but the sticky part, doesn't that alter the acceleration, and thus the speeD?

anonymous · Answer

yes, that's why it slowed over the distance given

anonymous · Answer

I hope that medal means you've gotten it. Regards, DWC

nottim · Answer

eh, got the wrong answer

anonymous · Answer

Sorry. Want to display your procedure? Or just move on?