Mark M. answered 07/25/19
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
a(t) = v'(t)
So, v'(t) = -6t - 18 and v(0) = 5
v(t) = ∫v'(t)dt = -3t2 - 18t + C
Since v(0) = 5, v(t) = -3t2 - 18t + 5
Rachel M.
asked 07/25/19An object with an initial velocity of 5 m/s undergoes an acceleration of a(t)=−6t−18 m/s^2, t seconds after an experiment begins.The velocity of the object after t seconds is v(t)=
Mark M. answered 07/25/19
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
a(t) = v'(t)
So, v'(t) = -6t - 18 and v(0) = 5
v(t) = ∫v'(t)dt = -3t2 - 18t + C
Since v(0) = 5, v(t) = -3t2 - 18t + 5
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