EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second).(a) Find the displacement of the particle during 1 ≤ t ≤ 6.(b) Find the distance traveled during this time period.SOLUTION(a) By this equation, the displacement iss(6) − s(1) = 6v(t) dt1 = 6(t2 − t − 20) dt1 = t33 − t22 − 20t 61 = −2756 .This means that the particle moved approximately 45.83 meters to the left.(b) Note that v(t) = t2 − t − 20 = (t − 5)(t + 4) and so v(t) 0 on the interval [1, 5] and v(t) 0 on [5, 6]. Thus, from this equation, the distance traveled is6|v(t)| dt1 = 5[−v(t)] dt1 + 6v(t) dt5 = 5(−t2 + t + 20) dt1 + 6(t2 − t − 20) dt5 = −t33+t22+20t 51 + t55+t22+20t 65 = 9.17 .
Question
EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second).(a) Find the displacement of the particle during 1 ≤ t ≤ 6.(b) Find the distance traveled during this time period.SOLUTION(a) By this equation, the displacement iss(6) − s(1) = 6v(t) dt1 = 6(t2 − t − 20) dt1 = t33 − t22 − 20t 61 = −2756 .This means that the particle moved approximately 45.83 meters to the left.(b) Note that v(t) = t2 − t − 20 = (t − 5)(t + 4) and so v(t) 0 on the interval [1, 5] and v(t) 0 on [5, 6]. Thus, from this equation, the distance traveled is6|v(t)| dt1 = 5[−v(t)] dt1 + 6v(t) dt5 = 5(−t2 + t + 20) dt1 + 6(t2 − t − 20) dt5 = −t33+t22+20t 51 + t55+t22+20t 65 = 9.17 .
Solution
The text you provided is in English, so I'll continue in English.
The problem you've given is a calculus problem involving the motion of a particle.
(a) The displacement of the particle from time t=1 to t=6 is calculated by integrating the velocity function v(t) = t^2 - t - 20 from 1 to 6. This gives the result -275/6, which means the particle has moved approximately 45.83 meters to the left.
(b) The distance traveled by the particle is the absolute value of the displacement. This is calculated by integrating the absolute value of the velocity function from 1 to 6. The velocity function changes sign at t=5, so the integral is split into two parts: from 1 to 5 and from 5 to 6. The result of these integrals is 9.17, which means the particle has traveled approximately 9.17 meters.
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