Math  /  Calculus

QuestionThe velocity function, in feet per second, is given for a particle moving along a straight line v(t)=t2t90,1t11v(t)=t^{2}-t-90,1 \leq t \leq 11 (a) Find the displacement (in feet). \square ft (b) Find the total distance (in feet) that the particle travels over the given interval.

Studdy Solution

STEP 1

1. The velocity function is v(t)=t2t90 v(t) = t^2 - t - 90 .
2. The time interval is from t=1 t = 1 to t=11 t = 11 .
3. Displacement is the net change in position, calculated as the definite integral of velocity over the given interval.
4. Total distance is the integral of the absolute value of velocity over the given interval.

STEP 2

1. Calculate the displacement by integrating the velocity function over the interval.
2. Determine when the velocity is zero to find intervals of positive and negative velocity.
3. Calculate the total distance by integrating the absolute value of the velocity function over the interval.

STEP 3

Calculate the displacement by integrating the velocity function from t=1 t = 1 to t=11 t = 11 .
Displacement=111v(t)dt=111(t2t90)dt \text{Displacement} = \int_{1}^{11} v(t) \, dt = \int_{1}^{11} (t^2 - t - 90) \, dt

STEP 4

Evaluate the integral:
(t2t90)dt=t33t2290t+C \int (t^2 - t - 90) \, dt = \frac{t^3}{3} - \frac{t^2}{2} - 90t + C
Calculate the definite integral:
[t33t2290t]111 \left[ \frac{t^3}{3} - \frac{t^2}{2} - 90t \right]_{1}^{11}

STEP 5

Substitute the limits of integration:
(1133112290×11)(13312290×1) \left( \frac{11^3}{3} - \frac{11^2}{2} - 90 \times 11 \right) - \left( \frac{1^3}{3} - \frac{1^2}{2} - 90 \times 1 \right)
Simplify the expression:
(133131212990)(131290) \left( \frac{1331}{3} - \frac{121}{2} - 990 \right) - \left( \frac{1}{3} - \frac{1}{2} - 90 \right)

STEP 6

Calculate the numerical values:
(443.6760.5990)(0.330.590) \left( 443.67 - 60.5 - 990 \right) - \left( 0.33 - 0.5 - 90 \right)
=(606.83)(90.17) = (-606.83) - (-90.17)
=606.83+90.17 = -606.83 + 90.17
=516.66 = -516.66
The displacement is:
516.66ft \boxed{-516.66} \, \text{ft}

STEP 7

Find when the velocity is zero to determine intervals of positive and negative velocity:
Solve v(t)=t2t90=0 v(t) = t^2 - t - 90 = 0 .

STEP 8

Use the quadratic formula to solve for t t :
t=b±b24ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1 a = 1 , b=1 b = -1 , c=90 c = -90 .
t=(1)±(1)241(90)21 t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-90)}}{2 \cdot 1}
t=1±1+3602 t = \frac{1 \pm \sqrt{1 + 360}}{2}
t=1±3612 t = \frac{1 \pm \sqrt{361}}{2}
t=1±192 t = \frac{1 \pm 19}{2}
t=10ort=9 t = 10 \quad \text{or} \quad t = -9
Since t=9 t = -9 is outside the interval, we only consider t=10 t = 10 .

STEP 9

Calculate the total distance by integrating the absolute value of the velocity function over the intervals [1,10] [1, 10] and [10,11] [10, 11] .
Total Distance=110v(t)dt+1011v(t)dt \text{Total Distance} = \int_{1}^{10} |v(t)| \, dt + \int_{10}^{11} |v(t)| \, dt

STEP 10

Since v(t)<0 v(t) < 0 for 1t<10 1 \leq t < 10 and v(t)>0 v(t) > 0 for 10<t11 10 < t \leq 11 , we have:
Total Distance=110v(t)dt+1011v(t)dt \text{Total Distance} = -\int_{1}^{10} v(t) \, dt + \int_{10}^{11} v(t) \, dt

STEP 11

Calculate the integrals:
110(t2t90)dt+1011(t2t90)dt -\int_{1}^{10} (t^2 - t - 90) \, dt + \int_{10}^{11} (t^2 - t - 90) \, dt
Evaluate each integral:
[t33t2290t]110+[t33t2290t]1011 -\left[ \frac{t^3}{3} - \frac{t^2}{2} - 90t \right]_{1}^{10} + \left[ \frac{t^3}{3} - \frac{t^2}{2} - 90t \right]_{10}^{11}

STEP 12

Substitute the limits of integration:
For [1,10] [1, 10] :
(1033102290×10)+(13312290×1) -\left( \frac{10^3}{3} - \frac{10^2}{2} - 90 \times 10 \right) + \left( \frac{1^3}{3} - \frac{1^2}{2} - 90 \times 1 \right)
For [10,11] [10, 11] :
(1133112290×11)(1033102290×10) \left( \frac{11^3}{3} - \frac{11^2}{2} - 90 \times 11 \right) - \left( \frac{10^3}{3} - \frac{10^2}{2} - 90 \times 10 \right)

STEP 13

Calculate the numerical values:
For [1,10] [1, 10] :
(1000350900)+(130.590) -\left( \frac{1000}{3} - 50 - 900 \right) + \left( \frac{1}{3} - 0.5 - 90 \right)
=(333.3350900)+(0.330.590) = -\left( 333.33 - 50 - 900 \right) + \left( 0.33 - 0.5 - 90 \right)
=(616.67)+(90.17) = -(-616.67) + (-90.17)
=616.6790.17 = 616.67 - 90.17
=526.5 = 526.5
For [10,11] [10, 11] :
(443.6760.5990)(333.3350900) \left( 443.67 - 60.5 - 990 \right) - \left( 333.33 - 50 - 900 \right)
=(606.83)(616.67) = (-606.83) - (-616.67)
=9.84 = 9.84
Total Distance:
526.5+9.84=536.34 526.5 + 9.84 = 536.34
The total distance is:
536.34ft \boxed{536.34} \, \text{ft}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord