SOLUTION: A subway train starts from rest at a station and accelerates at a rate of 1.66m/s^2 for 14.0s. It runs at constant speed for 70.9s and slows down at a rate of 3.50m/s^2 until it st

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Question 262863: A subway train starts from rest at a station and accelerates at a rate of 1.66m/s^2 for 14.0s. It runs at constant speed for 70.9s and slows down at a rate of 3.50m/s^2 until it stops at the next station. Find the total distance covered.
I know that there are 3 parts to this problem and to do them separatly and add up the distance at the end. I figured out the first part of the journey, being from rest through acceleration, (if I'm correct) using the formula , where v=initial velocity (o), a= acceleration (1.66m/s^2), and t= time (14.0s).

Found 2 solutions by Alan3354, palanisamy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A subway train starts from rest at a station and accelerates at a rate of 1.66m/s^2 for 14.0s. It runs at constant speed for 70.9s and slows down at a rate of 3.50m/s^2 until it stops at the next station. Find the total distance covered.
I know that there are 3 parts to this problem and to do them separatly and add up the distance at the end. I figured out the first part of the journey, being from rest through acceleration, (if I'm correct) using the formula , where v=initial velocity (o), a= acceleration (1.66m/s^2), and t= time (14.0s).
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The 1st part is s = at^2/2 = 0.83*196 meters (as you had) = 162.68 meters
It's speed is then at, = 1.66*14 m/sec
The 2nd part is vt = at*70.9 = 1.66*14*70.9 meters = 1647.716 meters
For the 3rd part, t = vt/3.5 = 1.66*14/3.5 = 6.64 seconds
The 3rd part is 1.66*14*6.64/2 = 77.1568 meters
total = 1887.5528

Answer by palanisamy(496)   (Show Source): You can put this solution on YOUR website!
First part of the journey with uniform acceleration
Initial velocity u = o
Let final velocity = v
Acceleration a= 1.66
Time taken t= 14 s
Then, v = u+at
v = 0+1.66*14 = 23.24 m/s ...(1)
Distance travelled x = ut+0.5at^2
x = 0*14+(0.5)*1.66*14*14
x = 162.68 m ...(2)
Second part of the journey with constant velocity
The train travels with a constant velocity of 23.24 m/s for 70.9 secs
Ditance travelled y = velocity*time
y = 23.24*70.9
y = 1647.16 m ...(3)
Third part of the journey with uniform retardration
Initial velocity u = 23.24
Final velocity v = 0
Acceleration a= -3.5
Let distance travelled = z
Then, v^2 = u^2+2as
0 = (23.24)^2-2*3.5*z
7z = 540.0976
z = 540.0976/7
z = 77.1568 m ...((4)
Total distance travelled = x+y+z
= 162.68+1647.16+77.1568
= 1886.9968 m

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