Respuesta :
Answer:
Acceleration = -12/17 m/s ^2
Explanation:
Name the variables
V = final velocity( velocity after the train stops) u= intial velocity(velocity when the train was moving at constant speed, before brakes are applied)
A = acceleration( in this case, it is deceleration)
T = time ( time taken to fully stop)
Second Step: What equation should we use?
The 3 Big Motion in One Dimension Formulas are:
V = u+at
v^2 = u^2 + 2as
S = ut+ (1/2) at^2
For this question, we will be using the first one.
Third Step: Solve
V = 0 (the final velocity is 0, because the train stopped)
U = 60 (the train was going at 60 mps)
T = 85 seconds( 1 min and 25 sec)
0 = 60+ 85a
First convert the minutes to seconds. That’s the standard unit for time in physics. 2 minutes is 120 seconds, add that to the 25 second to get 145 seconds.
Use the first kinematic equation Vf = Vi + at. We are given everything here besides acceleration. We know Vf equals zero because it says the train comes to a stop.
0 = 60m/s + a*145s
-60m/s = 145a
a = -60/145 = -12/29 m/s^2
Use the first kinematic equation Vf = Vi + at. We are given everything here besides acceleration. We know Vf equals zero because it says the train comes to a stop.
0 = 60m/s + a*145s
-60m/s = 145a
a = -60/145 = -12/29 m/s^2