Question

As a train accelerates away from a station, it reaches a speed of 5.2m/s in 5.0s. If the train's acceleration remains constant, what is the speed after an additional 6.0s has elapsed?

Answer 1

Does the problem states that the train started it motion from the rest ?

Answer 2

Nope.

Answer 3

But I assume that it is at rest when it was at the station.

Answer 4

then u can build up 3 equations using
v = u + at
where
v= final velocity
u = initial velocity
a = acceleration
t = time
1st one - the motion of train from the station until to the 5th sec
5.2 = u + 5a
hence u = 0
5.2 = 5a -----(1)

2nd one - the motion on train from 5th sec up to another 6 sec. onward;
final velocity = yet don't know
initial velocity = 5.2

v = u +at
v = 5.2 + 6a
6a = v - 5.2
v - 5.2 = 6a ------(2)

now divide (1)/(2) in a way to cancel out "a" and find out what's v.....
Hope this will help ya out!

Answer 5

\[\frac{ (1) }{ (2) } ---> \frac{ 5.2 }{ v -5.2 } = \frac{ 5a }{ 6a }\]
\[6*5.2 = 5*(v - 5.2)\]\
[31.2 = 5v -26\]
\[5v = 57.2\]
\[v =\frac{ 57.2 }{ 5 }\]
\[v = 11.44 ms ^{-1}\]

Answer 6

I'm still a bit confused about the concept. How do you cancel the accelerations to get the velocity if the accelerations are different? It says that the acceleration becomes 5.2 and remains constant, but what about the period of time before it reached 5.2m/s^2?

Answer 7

Yo... the question u post says "it reaches a speed of 5.2m/s in 5.0s" id doesn't says " acceleration becomes 5.2 and remains constant" ... Which one is correct bro.. ?

Answer 8

Should it be corrected like this ... ?
"it reaches a speed of 5.2m/s in 5.0s" to "it reaches an acceleration of 5.2m/s^2 in 5.0s

Answer 9

But it says "If the train's acceleration remains constant,"

Answer 10

no bro ... If the question u post is right. then the meaning of If "the train's acceleration remains constant" says that train traveled with the same constant acceleration even after the first five seconds. That phrase indicate that no change has happened to the acceleration in between 5th and 11th seconds of the motion

Answer 11

Ie would be a graph like this....