Answer to Question #251688 in Mechanics | Relativity for Kai

Question

a train moving with constant acceleration travels 24ft. during the 10th sec of its motion and 18 ft during the 12th sec of its motion.

a. find its initial velocity 

b. Compute the time required for the train to stop

c. How many feet will the travel before coming to rest?

Answer 1

a) The distance traveled during the first 10 and 11 seconds of its motion:

$d_{10}=v_0t-\frac{at_{10}^2}2=10v_0-50a,\\\space\\ d_{11}=v_0t-\frac{at_{11}^2}2=11v_0-60.5a,\\\space\\ d_{11}-d_{10}=24\text{ ft}. \\\space\\ 24=(11v_0-60.5a)-(10v_0-50a),\\ 24=v_0-10.5a.$

Now by analogy consider its motion between 12 and 13 seconds of its motion:

$d_{12}=v_0t-\frac{at_{12}^2}2=12v_0-72a,\\\space\\ d_{13}=v_0t-\frac{at_{13}^2}2=13v_0-84.5a,\\\space\\ d_{13}-d_{12}=18\text{ ft}. \\\space\\ 18=(12v_0-72a)-(13v_0-84.5a),\\ 18=12.5a-v_0.$

So, we have two equations:

$18=v_0-12.5a,\\ 24=v_0-10.5a.\\\space\\ a=3\text{ ft/s}^2,\\ v_0=55.5\text{ ft/s}.$

b) The time to stop is

$t=\frac va=18.5\text{ s}.$

c) The stopping distance is

$d=\frac{v^2}{2a}=513\text{ ft}.$