Question 825871
<pre>
Hi, there--

THE PROBLEM:
The equation {{{S=ut+(at)^2}}}  gives the distance S passed over by a moving body in t sec.
In 4 sec. the body moves 88 m.
In 6 sec. the body moves 168 m.

Find the values of u and a and then find how far the body moves in 5 s.

A SOLUTION:

By substituting the known values for u and a, we can find the exact equation.

When t = 4, S = 88.

{{{S=ut+(at)^2}}}
{{{88=u(4)+(a(4))^2}}}

Simplify.
{{{88=4u+16a^2}}}
{{{16a^2+4u=88}}}


When t = 6, S = 168
{{{S=ut+(at)^2}}}
{{{168=u(6) + (a(6))^2}}}

Simplify.
{{{168 = 6u + 36a^2}}}
{{{36a^2+6u=168}}}
{{{6a^2+u=28}}}

Now we have a system two equations with 2 variables. We can solve for u and a.

Rewrite the first equation in terms of u.
{{{16a^2+4u=88}}}
{{{4u=-16a^2+88}}}
{{{u=-4a^2+22}}}

Substitute {{{-4a^2+22}}} or u in the SECOND equation.
{{{6a^2+u=168}}} 
{{{6a^2+(-4a^t+22)=168}}}
{{{6a^2-4a^2+22=28}}}

Simplify.
(((2a^2+22=28}}}
{{{2a^2=6}}}
{{{a^2=3}}}
{{{a=sqrt(3)}}} or {{{a=-sqrt(3)}}}

Substitute these value for a into the "u=" equation.
{{{u=-4a^2+22}}}
{{{u=-4(sqrt(3))^2+22}}}
{{{u= -4(3)+22}}}
{{{u=10}}}

{{{a=-sqrt(3)}}} will give the same value for u because {{{(sqrt(3))^2=-(sqrt(3))^2}}}.

Now substitute the values for a and u into the original equation.

{{{S=ut+(at)^2}}}
{{{S=(10)t+((sqrt(3))^2*t^2)}}}
{{{S=10t+3t^2}}}

Find S when t=5.

{{{S=10(5)+3(5)^2}}}
{{{S=50+75}}}
{{{S=125}}}

In 5 sec. the body moves 125 m.

Hope this helps! Feel free to email if you have any questions.

Mrs. Figgy
math.in.the.vortex@gmail.com
</pre>