Linear Kinematics Question?

Question

Hi! I have a question that I need help on. I have been trying to figure it out for hours. A 87.0 m long train begins uniform acceleration from rest. The front of the train has a speed of 19.9 m/s when it passes a ralilway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answer 1

very very simple 24.234 m/s

here is how:
v=u+at 19.9=0+at
s=ut+1/2at^2180=1/2*at*t
180=1/2*19.9*t
t=18.09
at=19.9
a=1.1 m/s^2

v=u+at v=19.9+1.1*t
s=ut+1/2at^2 87=19.9*t+1/2*1.1*t*t
t=3.94
v=19.9+1.1*3.94
v=24.234
very basic school level physics

Answer 2

i try explain the problem in words , 1- front car started from rest u=0 acceleration unknown time also unknown (say t1) to pass ralilway worker (180m) with speed 19.9m/s
2- last car started with same speed of front car when passes ralilway worker (19.9m/s) with same accleration and different time (say t2). now the solution
for front car :
we use the relation v^2= u^2 + 2 a x so that the cceleratin a = v^2/2x where u =0 by subtitution a= 19.9^2/2*180 = 1.1m/s^2
for last car : here u= 19.9m/s x= 87m but v is uknown so that v^2=u^2 +2ax =19.9^ +2(1.1) (87) =396.01+ 191.4=587.41 so that the speed of last car square root of this value v=587.41^0.5=24.236 m/s note we can use t1 and t2 as Kaushal s doese but i think this easier .

Answer 3

I agree with the result kaushal s came up with but I think my choice of the basic motion formulas gives a simpler solution.

For the portion when the train is approaching the worker,
use V^2 = Vo^2 + 2as
(19.9 m/s)^2 = 2*a*180 m
a = 360/396 m/s^2 = 1.1 m/s^2

For the portion when the train is passing the worker,
use V^2 = Vo^2 + 2as again
V^2 = (19.9 m/s)^2 + 2*(1.1 m/s^)*87 m
v^2 = 396 m^2/s^2 + 191.4 m^2/s^2 = 587.4 m^2/s^2
So V = 24 m/s

Answer 4

65.3 m/s