i have a math/physics problem that i don't understand!?

Question

i didn't under stand the prof when he explained it.. please explain it to me(explain how do we get the answer) :
Velocity

Question: A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s} ^ {2} for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m}/s} ^ {2} until it stops at the next station. Find the total distance covered.

Answer: Acceleration phase:
s=1/2 at^2
s=1/2 * 1.6 *14^2 = 156.8 m

At that point, the train moves at v=at = 1.6 * 14=22.4 m/s, so for 70 seconds, the distance traveled at constant speed is 22.4 * 70 = 1568 m/

Deceleration phase:
v=a t
22.4 = 3.5 * t so t is 6.4 seconds
Distance traveled during deceleration is
s=1/2 a t^2 = 71.68 m

total distance traveled is thus 1796.48 m

Answer 1

Let us do the problem from first principles:

The train starts from rest and accelerates at 1.6 m/sec^2 for 14 secs. We need two things, one the velocity v that it reaches at the end of that 14 secs and the distance covered in that time of 14 secs.

v = u + at is the formula for the velocity.

s = ut + 1/2.a.t^2 is the formula for the distance

In our case, pl. remember that u = 0 since the train started from rest. So, we can get s and v by using the formulas.

s = 1/2 X 1.6 X 14 X 14 = 156.8 m and

v = 1.6 X 14 = 22.4 m/s

Now the train continues to move at this velocity for 70 secs and so the distance moved is simply velocity X time. Let us see why?

s = ut + 1/2at^2 is the formula where u is the initial velocity. Now the initial velocity is 22.4 m/s and acceleration is zero. Thus the second term is 0 and s = 22.4 X 70 = 1568 m

From that point, the velocity is reduced by applying brakes and the deceleration is given as 3.5 m/sec^2.

We can find the distance travelled s by

v = u - at or 0 = 22.4 - 3.5t or t = 22.4/3.5 = 6.4 seconds and

s = 22.4t - 1/2 X 3.5 X t^2

= 22.4 X 6.4 - 1/2 X 3.5 X 6.4 X 6.4 = 143.36 - 71.68 = 71.68m

The total distance travelled by the train from start to stop is the sum total of the three distances:

156.8 + 1568 + 71.68 = 1796.48 m

I will only correct the following portion:

Deceleration phase:
v=a t
22.4 = 3.5 * t so t is 6.4 seconds
Distance traveled during deceleration is
s=1/2 a t^2 = 71.68 m

The formula for s should be s = vt - 1/2.a.t^2 and not as used. In this case, the result happens to be same by both methods and your result is not affected. But the right formula is what I gave above.

Answer 2

HERE IS UR EXPLANATION
------------------------------------------
you are using the fundamental equation
s = ut + 1/2at^2
i hope u know wat all d letters stand for, but jus in case
s = distance
u = initial velocity
t = time of run
a = acceleration

i'm splitting the velocity,distance, time and accel as
u, s,t,a = accel phase
v, s',t',a' = constant velocity phase
v', s",t",a" = decel phase

PHASE I

initially, the velocity is zero
=> u = 0
time of acceleration is given
so use that t
this gives you
s = 0 + 1/2at^2

substituting values of a = 1.6, t = 14
s = 156.8

PAHSE I

now the body has accelerated for 14 seconds
hence using another fundamental equation
v = u + at
(v = final velocity)
In this case, u = 0
=> v = a*t = 22.4

distance travelled at this constant velocity for 70 s is
s' = 22.4 * t' = 1568 m

PHASE III

This is the most important phase
you can just calculate like ur teacher has shown, but the clearer method is as follows

in this you hav to use the final equation
v'^2 - v^2 = 2a''s''
here final velocity is zero
Hence
(22.4)^2 = 2*(-3.50)*(s'')

FINAL PART

solve this to find s''
s'' = 71.6

Hence, total distance =
s+s'+s" = 156.8 + 1568 + 71.68
= 1796.48

Answer 3

It should be clear to you if two cars are on the high way, traveling at different speeds, then one car will cover a further distance then the other car for the same traveling time.
The speed will determine how far the train travels in a given time.
Acceleration is a little like speed, but the speed of the acceleration is changing from moment to moment (rate of change of change). So depending on the acceleration value, will depend on the distance covered.
So basically, you have three parts to the question.
(1). You need to identify the distance covered with the initial acceleration (starting).
(2). You need to identify the distance covered with the constant speed part of the travel.
(3). You need to identify the distance covered with the final acceleration part of the travel (stopping).
Each part has covered a certain distance, so you add them all together to get the total distance.

Answer 4

All of this appears to be perfectly straightforward, and I don't see any problems with it. The proper formulas were properly applied and gave proper results.