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object's position given by s = 2t^2 - 3t; find time for....
- Thread starter freddie
- Start date
\(\displaystyle \L \;s\,={2t}^2\,-\,3t\)
\(\displaystyle s\,=\,length\,in\,meters\)
\(\displaystyle t\,=\,time\,in\,seconds\)
\(\displaystyle s\,=\,11\,meters\)
So solve:\(\displaystyle \L \;11\,=\,{2t}^2\,-\,3t\)
Once again, try using using the quadratic formula or completing the square. If you mess up we'll help you.
Re: solve the problem
Hello, freddie!
This is a trick question . . . a very sneaky one!
Since this is posted under "Algebra" (not Calculus), I'll explain in algebraic terms.
At the beginning: \(\displaystyle \,s(0)\:=\:2\cdot0^2\,-\,3\cdot0\:=\:0\)
. . The object is at the origin.
At \(\displaystyle t\,=\,1:\;s(1)\:=\:2\cdot1^2\,-\,3\cdot1\:=\:-1\)
. . The object is one unit to the left of the origin.
At \(\displaystyle t\,=\,2:\;s(2)\:=\:2\cdot2^2\,-\,3\cdot2\:=\:+2\)
. . The object is two units to the right of the origin.
Get it?The object changes direction!
Exactly when does it change direction?
The graph of \(\displaystyle s(t)\) is an up-opening parabola.
The minimum \(\displaystyle s\) occurs at its vertex.
The vertex is at: \(\displaystyle \,t\:=\:\frac{-b}{2a} \:=\:\frac{-(-3)}{2(2)} \:=\:\frac{3}{4}\)
And: \(\displaystyle \,s\left(\frac{3}{4}\right)\:=\:2\left(\frac{3}{4}\right)^2\,-\,3\left(\frac{3}{4}\right) \:=\:-\frac{9}{8}\)
That is, from \(\displaystyle t\,=\,0\) to \(\displaystyle t\,=\,\frac{3}{4}\), it moves \(\displaystyle \frac{9}{8}\,=\,1\frac{1}{8}\) units to the left.
Then it changes direction and heads toward the right another \(\displaystyle 9\frac{7}{8}\) units.
Since it started at \(\displaystyle \,-1\frac{1}{8}\), it must move to \(\displaystyle +8\frac{3}{4}\)
When is the object at \(\displaystyle 8\frac{3}{4}\,=\,\frac{35}{4}\) ?
. . When \(\displaystyle s(t)\:=\:2t^2\,-\,3t\:=\:\frac{35}{4}\)
Mutlply by 4 and we have: \(\displaystyle \,8t^2\,-\,12t\,-35\:=\:0\)
Quadratic Formula: \(\displaystyle \:t\:=\:\frac{-(-12)\,\pm\,\sqrt{(-12)^2\,-\,4(8)(-35)}}{2(8)} \;= \;\frac{3\,\pm\,\sqrt{79}}{4}\)
The positive root is: \(\displaystyle t \:=\:2.972048604 \:\approx\:\fbox{3.0\text{ seconds}}\)
That is, it takes about 3 seconds to move a total of 11 meters.
Hello, freddie!
This is a trick question . . . a very sneaky one!
The position of an object moving in a straight line is given by: \(\displaystyle \,s\:=\:2t^2\,-\,3t\)
where \(\displaystyle s\) is in meters and \(\displaystyle t\) is the time in second the object has been in motion.
How long (to the nearest tenth) will it take the object to move 11 meters?
Since this is posted under "Algebra" (not Calculus), I'll explain in algebraic terms.
At the beginning: \(\displaystyle \,s(0)\:=\:2\cdot0^2\,-\,3\cdot0\:=\:0\)
. . The object is at the origin.
At \(\displaystyle t\,=\,1:\;s(1)\:=\:2\cdot1^2\,-\,3\cdot1\:=\:-1\)
. . The object is one unit to the left of the origin.
At \(\displaystyle t\,=\,2:\;s(2)\:=\:2\cdot2^2\,-\,3\cdot2\:=\:+2\)
. . The object is two units to the right of the origin.
Get it?The object changes direction!
Exactly when does it change direction?
The graph of \(\displaystyle s(t)\) is an up-opening parabola.
The minimum \(\displaystyle s\) occurs at its vertex.
The vertex is at: \(\displaystyle \,t\:=\:\frac{-b}{2a} \:=\:\frac{-(-3)}{2(2)} \:=\:\frac{3}{4}\)
And: \(\displaystyle \,s\left(\frac{3}{4}\right)\:=\:2\left(\frac{3}{4}\right)^2\,-\,3\left(\frac{3}{4}\right) \:=\:-\frac{9}{8}\)
That is, from \(\displaystyle t\,=\,0\) to \(\displaystyle t\,=\,\frac{3}{4}\), it moves \(\displaystyle \frac{9}{8}\,=\,1\frac{1}{8}\) units to the left.
Then it changes direction and heads toward the right another \(\displaystyle 9\frac{7}{8}\) units.
Since it started at \(\displaystyle \,-1\frac{1}{8}\), it must move to \(\displaystyle +8\frac{3}{4}\)
When is the object at \(\displaystyle 8\frac{3}{4}\,=\,\frac{35}{4}\) ?
. . When \(\displaystyle s(t)\:=\:2t^2\,-\,3t\:=\:\frac{35}{4}\)
Mutlply by 4 and we have: \(\displaystyle \,8t^2\,-\,12t\,-35\:=\:0\)
Quadratic Formula: \(\displaystyle \:t\:=\:\frac{-(-12)\,\pm\,\sqrt{(-12)^2\,-\,4(8)(-35)}}{2(8)} \;= \;\frac{3\,\pm\,\sqrt{79}}{4}\)
The positive root is: \(\displaystyle t \:=\:2.972048604 \:\approx\:\fbox{3.0\text{ seconds}}\)
That is, it takes about 3 seconds to move a total of 11 meters.