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

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2as$where v = 23 m/s is the velocity of the car when it passes the worker, $v_0$ = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

$23^2 - 0^2 = 2*a*255$

$510a = 529$

$a = 529 / 510 = 1.04 m/s^2$