Question

A locomotive is accelerating at 3.08 m/s2. It passes through a 20.8-m-wide crossing in a time of 2.93 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 29.4 m/s?

Answer 1

Answer:

Required time t=5.8s

Explanation:

Given data

Acceleration a=3.08 m/s²

Speed v=29.4 m/s

Required

How much time t is required

Solution

As the train passes through the crossing its motion is described by:

$v=v_{o}+at\\v-v_{o}=at\\and\\x=\frac{1}{2}(v+v_{o})t\\v+v_{o}=\frac{2x}{t}\\ So\\v=\frac{1}{2}(at+\frac{2x}{t} )$

Substitute the given values

So

$v=\frac{1}{2}((3.08m/s^2)(2.93s)+\frac{2(20.8m)}{2.93s} )\\v=11.61m/s$

So time required can be calculated by:

$t=\frac{v-v_{o}}{a} \\t=\frac{29.4m/s-11.61m/s}{3.08m/s^2}\\ t=5.8s$

Required time t=5.8s