Math  /  Calculus

QuestionThe acceleration function (in m/s2\mathrm{m} / \mathrm{s}^{2} ) and the initial velocity v(0)v(0) are given for a particle moving along a line. a(t)=2t+4,v(0)=5,0t5a(t)=2 t+4, \quad v(0)=-5, \quad 0 \leq t \leq 5 (a) Find the velocity at time tt. v(t)=v(t)= \square m/s\mathrm{m} / \mathrm{s} (b) Find the distance traveled during the given time interval.

Studdy Solution
Calculate each integral:
01(t2+4t5)dt=[t332t2+5t]01 \int_{0}^{1} -(t^2 + 4t - 5) \, dt = \left[ -\frac{t^3}{3} - 2t^2 + 5t \right]_{0}^{1} =(132+5)(0) = \left( -\frac{1}{3} - 2 + 5 \right) - (0) =103 = \frac{10}{3}
15(t2+4t5)dt=[t33+2t25t]15 \int_{1}^{5} (t^2 + 4t - 5) \, dt = \left[ \frac{t^3}{3} + 2t^2 - 5t \right]_{1}^{5} =(1253+5025)(13+25) = \left( \frac{125}{3} + 50 - 25 \right) - \left( \frac{1}{3} + 2 - 5 \right) =1253+25(133) = \frac{125}{3} + 25 - \left( \frac{1}{3} - 3 \right) =1243+28 = \frac{124}{3} + 28 =1243+843 = \frac{124}{3} + \frac{84}{3} =2083 = \frac{208}{3}
The total distance is:
Distance=103+2083=2183 \text{Distance} = \frac{10}{3} + \frac{208}{3} = \frac{218}{3}
The velocity at time t t is:
v(t)=t2+4t5 v(t) = t^2 + 4t - 5
The distance traveled during the given time interval is:
2183 \boxed{\frac{218}{3}}

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