it travels from t=0 to t=4
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Distance = ∫|t - 3| dt from 0 to 4 = (3)(3)/2 + 1/2 = 5
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Integrate v = t - 3
s = (1/2) * t^2 - 3t + C
From t = 0 to t = 4
(1/2) * (4^2 - 0^2) - 3 * (4 - 0) + C - C =>
(1/2) * 16 - 12 =>
8 - 12 =>
-4
s = (1/2) * t^2 - 3t + C
From t = 0 to t = 4
(1/2) * (4^2 - 0^2) - 3 * (4 - 0) + C - C =>
(1/2) * 16 - 12 =>
8 - 12 =>
-4
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v = t - 3
dy/dt = t - 3
dy = (t - 3)dt
y = .5t^2 - 3t
Evaluating this integral from 0 to 4 yields:
y = .5(4)^2 - 3*4
y = 8 - 12
y = -4
dy/dt = t - 3
dy = (t - 3)dt
y = .5t^2 - 3t
Evaluating this integral from 0 to 4 yields:
y = .5(4)^2 - 3*4
y = 8 - 12
y = -4