Asked by Anonymous
The velocity of a car is recorded every 10 seconds, as shown in the table below.
t(sec) 0 10 20 30 40 50 60 70
v(t) (ft/sec) 4 12 22 38 50 57 59 60
A. Explain the meaning of integral[10,50] v(t) dt in terms of the car's travel. Use correct units.
B. Evaluate integral [20,60] v(t) dt using a left-hand approximation with n=4.
C. Evaluate integral [0, 60] v(t) dt using a midpoint approximation with n=3.
D. Evaluate integral [0, 70] v(t) dt using a trapezoidal approximation with n=7
E. Determine the average acceleration of the car over the time interval 0<=t<=70
t(sec) 0 10 20 30 40 50 60 70
v(t) (ft/sec) 4 12 22 38 50 57 59 60
A. Explain the meaning of integral[10,50] v(t) dt in terms of the car's travel. Use correct units.
B. Evaluate integral [20,60] v(t) dt using a left-hand approximation with n=4.
C. Evaluate integral [0, 60] v(t) dt using a midpoint approximation with n=3.
D. Evaluate integral [0, 70] v(t) dt using a trapezoidal approximation with n=7
E. Determine the average acceleration of the car over the time interval 0<=t<=70
Answers
Answered by
oobleck
A. ∫ v dt = ft/s * s = ft (so, total distance traveled)
B. ∫[10,50] v dt ≈ (v(20)+v(30)+v(40)+v(50))*10
B-D are similar
E. avg a = (v(70)-v(0))/(70-0) = 56/70
B. ∫[10,50] v dt ≈ (v(20)+v(30)+v(40)+v(50))*10
B-D are similar
E. avg a = (v(70)-v(0))/(70-0) = 56/70
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