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

Answer:

2621.25 meters

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, $x = v_{o}t+\frac{1}{2}at^{2}$

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

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