anonymous
  • anonymous
A particle moves along the x-axis with velocity v(t) = t^2 − 4, with t measured in seconds and v(t) measured in feet per second. Find the total distance travelled by the particle from t = 0 to t = 3 seconds.
Mathematics
jamiebookeater
  • jamiebookeater
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Astrophysics
  • Astrophysics
For the distance you need to know the positive and negative values because if you integrate from 0 to 3 you just have the displacement.
anonymous
  • anonymous
so how would you find those
Astrophysics
  • Astrophysics
Well we know v(t) = t^2-4 =(t+2)(t-2), notice this gives us the negative value when v(t) is going in the opposite directions, so you will have to take absolute values of when it goes from \[\left| \int\limits_{0}^{2} t^2-4 dt\right| + \left| \int\limits_{2}^{3} t^2-4dt \right|\] notice that distance is just \[\int\limits_{0}^{3} |v(t)| dt\]

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Astrophysics
  • Astrophysics
From 0
anonymous
  • anonymous
so the answer would be -3.0333?
Astrophysics
  • Astrophysics
No, remember we are taking the absolute value
Astrophysics
  • Astrophysics
That makes negative = +
anonymous
  • anonymous
oh yes so it would be 7.6
Astrophysics
  • Astrophysics
distance can't be negative, it's a scalar value
anonymous
  • anonymous
thank you so much!
Astrophysics
  • Astrophysics
Yes, that looks good \[\frac{ 23 }{ 2 } \approx 7.7\]
Astrophysics
  • Astrophysics
So it worked out?
anonymous
  • anonymous
yes it did and it made a lot of sense the way you explained it thank you

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