Question

A subway train starts from rest at a station and accelerates at a rate of 2.00 m/s2 for 14.0 s. It runs at constant speed for 66.0 s and slows down at a rate of 4.00 m/s2 until it stops at the next station. Find the total distance covered.

Answer 1

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 2 X 14 X 14 = 196 m and

v = 2 X 14 = 28 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 28 m/s and acceleration is zero. Thus the second term is 0 and s = 28 X 66 = 1452 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 = 28 - 4t or t = 28/4 = 7 seconds and

s = 28t - 1/2 X 4 X t^2

= 28 X 7 - 1/2 X 34 X 7 X 7

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.