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
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
A. The integral[10,50] v(t) dt represents the area under the velocity-time curve between the times 10 seconds and 50 seconds. In terms of the car's travel, it represents the total displacement or distance covered by the car between those two time points. Since velocity is measured in ft/sec and time is measured in seconds, the unit of the integral will be ft.
B. To evaluate the integral [20,60] v(t) dt using a left-hand approximation with n=4, we divide the interval [20,60] into 4 sub-intervals of equal width:
20 to 30
30 to 40
40 to 50
50 to 60
Using the left endpoint of each sub-interval to approximate the velocity at that time, we get:
v(20) = 22 ft/sec
v(30) = 38 ft/sec
v(40) = 50 ft/sec
v(50) = 59 ft/sec
The width of each sub-interval is 10 seconds. Using the left-hand approximation formula:
integral [20,60] v(t) dt ≈ 10*(v(20) + v(30) + v(40) + v(50))
= 10*(22 + 38 + 50 + 59)
= 10*169
= 1690 ft
So, the approximate value of the integral [20,60] v(t) dt is 1690 ft.
C. To evaluate the integral [0,60] v(t) dt using a midpoint approximation with n=3, we divide the interval [0,60] into 3 sub-intervals of equal width:
0 to 20
20 to 40
40 to 60
Using the midpoint of each sub-interval to approximate the velocity at that time, we get:
v(10) = 12 ft/sec
v(30) = 38 ft/sec
v(50) = 59 ft/sec
The width of each sub-interval is 20 seconds. Using the midpoint approximation formula:
integral [0,60] v(t) dt ≈ 20*(v(10) + v(30) + v(50))
= 20*(12 + 38 + 59)
= 20*109
= 2180 ft
So, the approximate value of the integral [0,60] v(t) dt is 2180 ft.
D. To evaluate the integral [0,70] v(t) dt using a trapezoidal approximation with n=7, we divide the interval [0,70] into 7 sub-intervals of equal width:
0 to 10
10 to 20
20 to 30
30 to 40
40 to 50
50 to 60
60 to 70
Using the trapezoidal approximation formula:
integral [0,70] v(t) dt ≈ 5*((v(0) + v(10))/2 + (v(10) + v(20))/2 + (v(20) + v(30))/2 + (v(30) + v(40))/2 + (v(40) + v(50))/2 + (v(50) + v(60))/2 + (v(60) + v(70))/2)
= 5*((4 + 12)/2 + (12 + 22)/2 + (22 + 38)/2 + (38 + 50)/2 + (50 + 57)/2 + (57 + 59)/2 + (59 + 60)/2)
= 5*(8 + 17 + 30 + 44 + 53.5 + 58 + 59.5)
= 5*(270.5)
= 1352.5 ft
So, the approximate value of the integral [0,70] v(t) dt is 1352.5 ft.
E. To determine the average acceleration of the car over the time interval 0 <= t <= 70, we can use the formula:
Average Acceleration = (Change in Velocity) / (Change in Time)
The change in velocity is given by v(70) - v(0) = 60 ft/sec - 4 ft/sec = 56 ft/sec.
The change in time is given by 70 sec - 0 sec = 70 sec.
Average Acceleration = 56 ft/sec / 70 sec = 0.8 ft/sec^2
So, the average acceleration of the car over the time interval 0 <= t <= 70 is 0.8 ft/sec^2.