Physics Train Problem?

Question

A 75m long train accelerates uniformly from rest. If the front of the train passes a railway worker 140m down the track at a speed of 25m/s, what will be the speed of the last car as it passes the worker?

I really just don't understand how to work this problem out. I'm not even sure where to start. If someone could give me a hand, I would greatly appreciate it.

Answer 1

Front of train travels 140m

Back of train travels 140 +75 = 215 m

We need to carry out calculations for front and back separately

U = initial velocity = 0 for both front and back

V = final velocity Vfront = Vf  = 25m/s and Vback = Vb

S= distance travelled Sfront = Sf = 140m and Sback = Sb = 215m

A = Acceleration same for both

Use numbers for front to calculate Acceleration A

Using general formula V^2 = U^2 + 2 A S substitute values for front

25 x 25 = 0 + 2 x A x 140

A = (25 x 25)/280 = 2.23 m/s^2 to 2 dec places

Now solve for Vb using U, A and Sb which is 215m

Vb^2 = U^2 + (2 x 2.23 x 215)

Vb = √ (0 + 2 x 2.23 x 215) = √ 958.9

Velocity of back of train = √ 958.9 = 30.97 m/s

Answer 2

As the train has a 140-m displacement, its velocity change from v0 = 0 to v1 = 25 m/s, the acceleration can be deduced from this formula:

v1² - v0²= 2as

a=  (v1² - v0²)/2s = (25² - 0²)/2×140

a = 625/280 = 125/56 m/s²

As the train moves 75 meter further, its velocity will have new value v2, defined by

v2² - v1² = 2as'

v2 = sqrt(v1² + 2as')

v2 = sqrt(25² + 2×75×125/56) = sqrt(625 + 18750/56)

v2 = 31m/s