Answer to Question #94783 in Mechanics | Relativity for June Vallery Del Rosario

Question

A body from rest is accelerated uniformly to a velocity of 12.25m/s in 15.0s. Then it continues to move at this velocity for an additional 20.0s. Then it slows down within 25.0s and finally stops. Find the acceleration during the first 15.0s and during the last 25.0s. And compute the total displacement of the body assuming that it is moving in a straight line.

Answer 1

Solution:

Let's find the acceleration during the first 15.0 s:

$a_1=\frac{\upsilon_1-\upsilon_0}{t_1}=\frac{12.25-0}{15.0}=0.82$ m/s².

Let's find the acceleration during the last 25.0 s:

$a_2=\frac{\upsilon_2-\upsilon_1}{t_3}=\frac{0-12.25}{25.0}=-0.49$ m/s².

Let's find the total displacement of the body:

$S=S_1+S_2+S_3=\frac{|a_1|t^2_1}{2}+\upsilon_1t_2+\frac{|a_2|t^2_3}{2}=$

$=\frac{0.82\cdot{(15.0)^2}}{2}+12.25\cdot{20.0}+\frac{0.49\cdot{(25.0)^2}}{2}=490.38$ m.

Answer:

a1=0.82 m/s², a2=-0.49 m/s², S=490.38 m.