Question

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

The displacement of the train in this time period is 2,616.86 m.

Explanation:

A Uniformly Varied Rectilinear Motion is Rectilinear because the mobile moves in a straight line, Uniformly because of there is a magnitude that remains constant (in this case the acceleration) and Varied because the speed varies, the final speed being different from the initial one.

In other words, a motion is uniformly varied rectilinear when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time (the speed is constant and the acceleration is variable).

An independent equation of useful time in this type of movement is:

$vf^{2} =vi^{2} +2*a*d$ Expression A

where:

• vf = final velocity
• vi = initial velocity
• a = acceleration
• d = distance

The equation of velocity as a function of time in this type of movement is:

vf=vi + a*t

So the velocity can be calculated as: $a=\frac{vf-vi}{t}$

In this case:

• vf=42.4 m/s
• vi=27.5 m/s
• t=75 s

Replacing in the definition of acceleration:  $a=\frac{42.4 m/s-27.5 m/s}{75 s}$

a=0.199 m/s²

Now, replacing in expression A:

$(42.4 m/s)^{2} =(27.5 m/s)^{2} +2*(0.199 m/s^{2}) *d$

Solving:

$d= \frac{(42.4 m/s)^{2} - (27.5 m/s)^{2} }{2*(0.199 m/s^{2}) }$

d= 2,616.86 m

The displacement of the train in this time period is 2,616.86 m.