The Speed Of A Train Is Reduced From 80km/h To 40km/h In A Distance Of 500m On Applying The Brakes (I)how Much Further Will The Train Travel Before Coming To Rest (II)assuming The Retardation Remain Constant,how Long Will It Take To Bring The Train To Rest After The Application Of The Brakes
average speed during initial slowdown = (80+40)/2 = 60km/h
60 km/h * t = 0.5 km
t = 5/600 hours = (1/120)hr
(1 hr/120)(60min/hr)(60s/min)
= 30 seconds
so
a = change in speed/time
= -40 km/hr /30 s
= (-40 * 1000 m/3600s)/30s
= -(4/3)/3.6 m/s^2
so
Vi = (40/3.6) m/s
final v = 0 =Vi+at
time to stop = -Vi/a
= (40/3.6) / [(4/3)/3.6]
=10*3 seconds
=30 seconds more
distance in final phase =(20/3.6)m/s(30s) =600/3.6 = 167 m more
total time = 30+30 = 60 seconds
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To calculate the answer, we need to apply the formula for acceleration:
1. (I) How much further will the train travel before coming to rest?
First, we need to calculate the deceleration (retardation) of the train.
Given:
Initial speed (u) = 80 km/h
Final speed (v) = 40 km/h
Distance (s) = 500 m
To convert km/h to m/s, we need to multiply by 1000/3600 (since 1 km/h = 1000 m / 3600 s).
Initial speed (u) = 80 km/h * (1000 m/3600 s) = 22.22 m/s
Final speed (v) = 40 km/h * (1000 m/3600 s) = 11.11 m/s
Using the formula:
v^2 = u^2 + 2as
Rearranging the formula to solve for distance (s):
s = (v^2 - u^2) / (2a)
where:
s is the distance traveled,
u is the initial velocity,
v is the final velocity, and
a is the acceleration (negative because it's deceleration).
Substituting the known values:
s = (11.11^2 - 22.22^2) / (2a)
To find a, we use the formula for acceleration:
a = (v - u) / t
where:
t is the time taken.
Since the train comes to rest, the final velocity (v) is 0 m/s.
Rearranging the formula to solve for time (t):
t = (v - u) / a
Substituting the known values:
0 = (11.11 - 22.22) / a
Simplifying the equation:
11.11 = 22.22 / a
From this equation, we can solve for a:
a = 22.22 / 11.11 = 2 m/s^2
Substituting the value of a back into the formula for distance (s):
s = (11.11^2 - 22.22^2) / (2 * 2)
Calculate s:
s = (123.21 - 493.04) / 4
s = -369.83 / 4
s = -92.46 m
The distance traveled before coming to rest is 92.46 meters.
2. (II) Assuming the retardation remains constant, how long will it take to bring the train to rest after the application of the brakes?
We already know the deceleration (retardation) a = 2 m/s^2 and the initial velocity u = 22.22 m/s.
Using the formula:
t = (v - u) / a
Substituting the values:
t = (0 - 22.22) / 2
Calculate t:
t = -22.22 / 2
t = -11.11 s
Since the time can't be negative, we take the absolute value:
t = 11.11 s
It will take approximately 11.11 seconds to bring the train to rest after the application of the brakes.