Velocity Functions

[azn1x6flame]

Hello,

I'm trying to do the problem below:

The velocity function (in meters per second) is given for a particle moving along a line,

v(t)=3t-11, 0 (greater than or equal to) t (greater than or equal to) 5

Find the distance (in meters) traveled by the particle during the given time interval.


I attempted the problem with this: 3(5)-11=4

But my answer is wrong, could you give pointers on how to solve this problem?

Thanks.

 

[tkhunny]

Did you use any calculus?

0 <= t <= 5

You should know that an antiderivative of v(t) will tell you where a partical is.

 

[azn1x6flame]

Did you use any calculus?

0 <= t <= 5

You should know that an antiderivative of v(t) will tell you where a partical is.

How would you find t?

Do you have do the derivative of v(t) to find acceleration? And would that help me to find t?

 

[soroban]

Hello, azn1x6flame!

This is a Calculus problem.
Did you expect to solve it with a little Arithmrtic?

\(\displaystyle \text{The velocity function (in m/sec) for a particle moving along a line is: }\:v(t)\:=\:3t-11,\; 0 \le t \le 5\)

\(\displaystyle \text{Find the distance traveled by the particle during the given time interval.}\)

\(\displaystyle \text{The distance function is: }\;s(t) \;=\;\int(3t-11)\,dt \;=\;\tfrac{3}{2}t^2 - 11t + C\)

The following is the wrong answer.

\(\displaystyle \text{At }t = 0\text{, the particle is at: }\:s(0) \:=\:\tfrac{3}{2}(0^2) - 11(0) + C \:=\:C\)

\(\displaystyle \text{At }t = 5\text{, the particle is at: }\:s(5) \:=\:\tfrac{3}{2}(5^2) - 11(5) + C \:=\:-\tfrac{35}{2} + C\)

\(\displaystyle \text{Therefore, the particle moved: }\-\tfrac{35}{2} \,+\, C) \,-\, C \:=\:-\tfrac{35}{2} \quad\Rightarrow\quad 17\tfrac{1}{2}\text{ units to the left.}\) . Wrong!


Does the particle ever stop and reverse direction?
\(\displaystyle \text{This happens when }v(t) = 0.\)
. . \(\displaystyle 3t - 11 \:=\:0 \quad\Rightarrow\quad t \:=\:\tfrac{11}{3}\)

\(\displaystyle \text{At }t = 0\text{, the particle is at: }\:s(0) \:=\:\tfrac{3}{2}(0^2) - 11(0) + C \:=\:C\)


\(\displaystyle \text{At }t = \tfrac{11}{3}\text{, the particle is at: }\:s(\tfrac{11}{3}) \:=\:\tfrac{3}{2}(\tfrac{11}{3})^2 - 11(\tfrac{11}{3}) + C \:=\:-\tfrac{121}{3} + C\)

. . \(\displaystyle \text{It has moved: }\-\tfrac{121}{3} + C) - C \:=\:-\tfrac{121}{3}\;\hdots\;\text{ It moved }40\tfrac{1}{3}\text{ units to the left.}\)


\(\displaystyle \text{At }t = 5\text{, the particle is at: }\:s(5) \:=\:\tfrac{3}{2}(5^2) - 11(5) + C \:=\:-\tfrac{35}{2} + C\)

. . \(\displaystyle \text{It has moved: }\-\tfrac{35}{2} + C) - (-\tfrac{121}{3} + C) \:=\:\tfrac{137}{6}\;\hdots\;\text{ It moved }22\tfrac{5}{6}\text{ units to the right.}\)


\(\displaystyle \text{Therefore, it moved a total distance of: }\:40\tfrac{1}{3} + 22\tfrac{5}{6} \;=\;63\tfrac{1}{6}\text{ units.}\)

 

[azn1x6flame]

\(\displaystyle \text{Therefore, it moved a total distance of: }\:40\tfrac{1}{3} + 22\tfrac{5}{6} \;=\;63\tfrac{1}{6}\text{ units.}\)

I was getting -17.5, but what you said makes sense.

Thanks soroban!

 

[tkhunny]

If only you had responded better to my post. You may have figured it out for yourself and gotten a warm fuzzy feeling about your personal accomplishment.