Math  /  Calculus

Question=autosave\#questio... ns that the particle moved approximately 30.00 meters to the left. that v(t)=t2t20=(t5)(t+4)v(t)=t^{2}-t-20=(t-5)(t+4) and so v(t)v0v(t) \leq v \quad 0 on the interval [1,5][1,5] and v(t)v(t) nus, from this equation, the distance traveled is 17v(t)dt=15[v(t)]dt+57v(t)dt=15(t2+t+20)dt+57(t2t20)dt\begin{aligned} \int_{1}^{7}|v(t)| d t & =\int_{1}^{5}[-v(t)] d t+\int_{5}^{7} v(t) d t \\ & =\int_{1}^{5}\left(-t^{2}+t+20\right) d t+\int_{5}^{7}\left(t^{2}-t-20\right) d t \end{aligned} =[t33+t22+20t]15+[t33t2220t]=\left[-\frac{t^{3}}{3}+\frac{t^{2}}{2}+20 t\right]_{1}^{5}+\left[\frac{t^{3}}{3}-\frac{t^{2}}{2}-20 t\right] =7253=\frac{725}{3} Your answer is correct.

Studdy Solution

STEP 1

1. The velocity function is given by v(t)=t2t20 v(t) = t^2 - t - 20 .
2. The function is factored as v(t)=(t5)(t+4) v(t) = (t-5)(t+4) .
3. The velocity is non-positive on the interval [1,5][1, 5] and changes sign at t=5 t = 5 .
4. We need to find the total distance traveled from t=1 t = 1 to t=7 t = 7 .

STEP 2

1. Determine the intervals where the velocity is negative and positive.
2. Set up the integral for the total distance traveled using the absolute value of the velocity.
3. Evaluate the integral over each interval.
4. Sum the results to find the total distance.

STEP 3

Determine the intervals where v(t) v(t) is negative and positive: - The roots of v(t)=0 v(t) = 0 are t=5 t = 5 and t=4 t = -4 . - v(t)0 v(t) \leq 0 on [1,5][1, 5] and v(t)0 v(t) \geq 0 on [5,7][5, 7].

STEP 4

Set up the integral for the total distance traveled:
17v(t)dt=15[v(t)]dt+57v(t)dt \int_{1}^{7} |v(t)| \, dt = \int_{1}^{5} [-v(t)] \, dt + \int_{5}^{7} v(t) \, dt

STEP 5

Evaluate the integral over the interval [1,5][1, 5]:
15(t2+t+20)dt \int_{1}^{5} (-t^2 + t + 20) \, dt
Calculate the antiderivative:
[t33+t22+20t]15 \left[-\frac{t^3}{3} + \frac{t^2}{2} + 20t\right]_{1}^{5}

STEP 6

Evaluate the antiderivative from t=1 t = 1 to t=5 t = 5 :
(533+522+20×5)(133+122+20×1) \left(-\frac{5^3}{3} + \frac{5^2}{2} + 20 \times 5\right) - \left(-\frac{1^3}{3} + \frac{1^2}{2} + 20 \times 1\right)

STEP 7

Evaluate the integral over the interval [5,7][5, 7]:
57(t2t20)dt \int_{5}^{7} (t^2 - t - 20) \, dt
Calculate the antiderivative:
[t33t2220t]57 \left[\frac{t^3}{3} - \frac{t^2}{2} - 20t\right]_{5}^{7}

STEP 8

Evaluate the antiderivative from t=5 t = 5 to t=7 t = 7 :
(73372220×7)(53352220×5) \left(\frac{7^3}{3} - \frac{7^2}{2} - 20 \times 7\right) - \left(\frac{5^3}{3} - \frac{5^2}{2} - 20 \times 5\right)

STEP 9

Sum the results from the two intervals to find the total distance:
The result of the integration is:
7253 \frac{725}{3}
The total distance traveled by the particle is:
7253 \boxed{\frac{725}{3}}

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