Question

A train traveling at 27.5 m/s accelerates to 42.4 m/s over 75.0 s. What is the displacement of the train in this time period

Answer 1

2621.25 meters

Step-by-step explanation:

First, write down what we are given.

Initial velocity = 27.5 m/s

Final velocity = 42.4 m/s

Time = 75 seconds

We need to look at the kinematic equations and determine which one will be best. In this case, we need an equation with distance. I am going to use
`v_(f)^(2) = v_(i)^(2) +2ad`, but you can also use the other equation,
`x = v_(o)t+(1)/(2)at^(2)`

We need to find acceleration. To find it, we need to use the formula for acceleration:
`a = (v_(f)-v_(i))/(t)`. Plugging in values,
`a = (42.4-27.5)/(75) = .199\ m/s^(2)`

Next, plug in what we know into the kinematics equation and solve for distance.
`42.4^(2) = 27.5^(2) + 2(.199)(d)\\d = 2621.25\ meters`