Question

A train is moving along a straight track with an initial velocity of 8 m/s when it is subjected to a deceleration of a-1.6/v m/s2, where v is in m/s. 2. How far does the train travel before it stops? How much time does it take to stop? a. b.

Answer 1

To calculate a.) we will use the formula

v² - u² = 2*a*s where v is final velocity, u is initial velocity and s is the distance.

For b.) we will use the formula

v = u +at

Given initial velocity, u = 8 m/s Final velocity v = 0 m/s deceleration a = -1.6 squareroot v m/s^2 v^2 - u^2 = 2 a s: s - distance 0 - (8)^2 = 2 (- 1.6 squareroot v) s - 64 = -2 times 1.6 squareroot v s s = 64/2 times 1.6 squareroot v = 32 times 10/16 squareroot v = 20/squareroot v therefore s = 20/squareroot v m implies Required distance v = u + a t 0 = 8 + (- 1.6 squareroot v) t 1.6 squareroot v t = 8 implies t = 8/1.6 squareroot v t = 8 times 10/16 squareroot v = 5/squareroot v seconds t = 5 s/squareroot v implies Required time