1. A train traveling at 20 m/s
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Break it down:
Initial speed = 20m/s. if train moves along +x axis, then velocity vector is Vi = + 20m/s or 20m/s east
A train traveling at 20 m/s
Deceleration means that the acceleration vector is in opposite direction of velocity vector. Therefore speed decreases or Vf < Vi
decelerates
Magnitude of acceleration vector. a = 2.0 m/s2 or a = - 2 m/s2 or 2 m/s2 west.
at 2.0 m/s2
If train stops speed at final instant will be zero Vf = 0
to stop at the station.
t =? , time interval between initial and final instants
(a) How much time does it take for the train to reach the station?
Distance between initial and final instants or time interval from slowing down to stopping (entire trip).
(b) How far
is train from the station when it starts slowing down
(c) How far does it travel during the fifth second
What is the distance between fifth and sixth instants? “during fifth second” means between seconds five and six. In other words for this part of problem initial time is second five and final time is second six. Then time interval for this part will be 6 – 5 = 1 second, or simply ∆t = 1sec.
after the brakes are applied?
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2. A train traveling at 20 m/s decelerates at 2
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Solution:
Draw a diagram
Vi = 20m/s
Vf = 0.0
a = 2m/s2, a = - 2 m/s2
t =?
X =? f i
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3. Use appropriate equation to solve for unknown
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Solution:
Use appropriate equation to solve for unknown
(a) How much time does it take for the train to reach the station?
Vf = at + Vi 
0 = (- 2m/s2 * t )+ (+20 m/s)  m/s = (- 2 m/s2 * t)
 t = - 20m/s/-2m/s2  t = 10s
When a variable moves to the other side of equal sign its sing will be changed.
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4. Use appropriate equation to solve for unknown
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Solution:
Use appropriate equation to solve for unknown
(b) How far is train from the station when it starts slowing down?
X= ½ (at 2) + Vi t
 X = ½ [(-2 m/s2) ( 10s) 2] + ( 20m/s * 10s)
 X = ½ [-200 m ] + 200m  X = + 100m
+ Sign shows that train moves 100m to east before it stops.
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5. ∆t = 6s – 5s = 1s, duration of time interval.
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Solution:
Use appropriate equation to solve for unknown
(c) How far does it travel during the fifth second after the brakes are applied?
We draw diagram here with more details
∆t = 6s – 5s = 1s, duration of time interval.
We have to find the traveled distance (∆X = X6 – X5) in this one second.
We can solve this part in two ways.
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6. the whole trip, including this interval
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
First solution:
find speeds in instants fifth and sixth (V5 and V6),
Then use V6 2 – V5 2 = 2 a ∆x
We know that acceleration is constant during
the whole trip, including this interval
Calculate V5, speed after 5 seconds from initial instant;
V5 - Vi = at
 V5 – (+20) = - 2*5 V5= -10 m/s + 20 m/s = +10m/s
V5 is speed after 5 seconds after initial instant that trains starts slowing down.
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7. Travels 9 m east from instant 5 to instant 6
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
First solution:
Calculate V6, speed after 6 seconds from initial instant;
V6 – Vi = at
 V6 – (+20 m/s) = (-2.0 m/s 2) * 6s
 V6 = - 12m/s + 20 m/s V6 = + 8 m/s
Now use time independent equation to find ∆x:
V62 – V52 = 2a ∆x
 [V6 2 – V5 2 ] / 2a = ∆x
∆x = [(+8)2 – (+10) 2] / [2(-2)] = [ ]/ (-4) = -36/ -4  ∆x = +9m
Travels 9 m east from instant 5 to instant 6
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8. We do not need to find V6. We can use ∆x =1/2[a (∆x) 2] + [V5 * ∆t ]
A train traveling at 20 m/s decelerates at 2.0 m/s2 to stop at the station. (a) How much time does it take for the train to reach the station? (b) How far is train from the station when it starts slowing down (c) How far does it travel during the fifth second after the brakes are applied?
Second solution:
We do not need to find V6. We can use ∆x =1/2[a (∆x) 2] + [V5 * ∆t ]
Remember, ∆t = 1 s, time interval between instant 5 & instant 6
∆x = ½[(-2m/s) (1s)2] + [(+10 m/s) * 1s] = (-1)m + 10 m
∆x = -1m + 10m = + 9m or 9m east
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