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[SOLVED]Constant Acceleration problem
When a high speed train traveling at 161 km/hr rounds a bend,he is shocked to see that a locomotive has improperly entered onto the track at a distance D=676 m ahead. The locomotive is moving at 29.0 km/hr away from the train. The engineer of the train immediately applies the brakes. What must be the magnitude of the resulting deceleration if a collision is just to be avoided?
a=C
v= vi+a*t
x=xi+vt+(1/2)*a*t^2
First i convert the speed of the train and the locomotive to m/s:
train: 44.7 m/s
Loco: .806 m/s
Next I make a system of eq. for x
x= (0)+(44.7)*t+.5*a*t^2 [kinematic equation]
x= (676) + .806*t {Ini. spacing add the distance trav. by Loc}
Now I combine the two
676= 43.9*t+.5*a*t^2
next I solve for a with respect to t
0 = 44.7+a*t [kinematic equation]
a = -44.7/t
and substitute and solve for t
676= 43.9*t - .5*(44.7*t)
t= 31.3
now I solve for a
a= -44.7/(31.3) [kinematic]
a= -1.42 m/sec^2 ANSWER
However the book gives something different (.994 m/sec^2), I think I'm screwing up in determining the time. Can someone help me in determining my error. Thanks in advance
edit: to make it less confusing
Homework Statement
When a high speed train traveling at 161 km/hr rounds a bend,he is shocked to see that a locomotive has improperly entered onto the track at a distance D=676 m ahead. The locomotive is moving at 29.0 km/hr away from the train. The engineer of the train immediately applies the brakes. What must be the magnitude of the resulting deceleration if a collision is just to be avoided?
Homework Equations
a=C
v= vi+a*t
x=xi+vt+(1/2)*a*t^2
The Attempt at a Solution
First i convert the speed of the train and the locomotive to m/s:
train: 44.7 m/s
Loco: .806 m/s
Next I make a system of eq. for x
x= (0)+(44.7)*t+.5*a*t^2 [kinematic equation]
x= (676) + .806*t {Ini. spacing add the distance trav. by Loc}
Now I combine the two
676= 43.9*t+.5*a*t^2
next I solve for a with respect to t
0 = 44.7+a*t [kinematic equation]
a = -44.7/t
and substitute and solve for t
676= 43.9*t - .5*(44.7*t)
t= 31.3
now I solve for a
a= -44.7/(31.3) [kinematic]
a= -1.42 m/sec^2 ANSWER
However the book gives something different (.994 m/sec^2), I think I'm screwing up in determining the time. Can someone help me in determining my error. Thanks in advance
edit: to make it less confusing
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