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
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}

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