Question

a 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tanding 180 m from where the front of the train started what will be the speed of the last car as it passes the worker? front of moving. what will 75 m y23 m/s

Answer 1

Train length = 75m

Initial velocity = 0

A worker is 180 m from the front of the train.

The front passes the worker going at 23 m/s.

The rear end is 75 + 180m from the worker = 255 m

We need to find the acceleration :

We will use the following equations of motion.

V = u + at

But u = 0

Doing the substitution we have :

23 = at

a = 23/t

S = u + 0.5at²

Again u = 0

180 m = 0.5at²

a = 180/0.5t²

a = 360/t²

360/t² = 23/t

360t = 23t²

Divide through by 23t we have :

360t/23t = t

t = 15.65 seconds.

a = 23/15.65 = 1.4696 m/s²

Now we can calculate velocity of last car.

S = 0.5at² since u =0

255 = 0.5 × 1.4696 × t²

t² = 255/0.7348

t² = 347.03

t = √347.03

t = 18.63 seconds.

V = at

= 18.63 × 1.4696 = 27.38 m/s

= 27.38 m/s