Question

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one wheel begins to stick. The train comes to a stop 7.7 m from the point at which it was released.
What is the train's acceleration after its wheel begins to stick?

Answer 1

I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

Answer 2

Answer:

$a = -1.16m/s^2$

Explanation:

Given

Initial speed u = 2.5 m/s

Final speed v = 0

Distance traveled S = 7.7 m

Duration t = 2.0 s

Solution

Distance traveled before the wheel got stuck

$d = ut\\\\d= 2.5 \times 2\\\\d = 5 m$

Distance traveled after wheel got stuck

$S'  = S - d\\\\S' = 7.7-5\\\\S' = 2.7 m$

Acceleration

$v^2 = u^2 + 2as'\\\\0 = 2.5^2 + 2 \times a times 2.7\\\\a = -1.16m/s^2$