M14056009 Key Questions
1. 7.5/8
The height in metres of a ball dropped from the top of the CN Tower is given by h(t)= -4.9t2+450, where t is time elapsed in seconds.
(a) Draw the graph of h with respect to time
(b) Find the average velocity for the first 2 seconds after the ball was dropped h(0)=(0,450), h(2)=(2,430.4)
= (430.4-450)/(2-0)
= -9.8m/s √
(c) Find the average velocity for the following time intervals
(1) 1 ≤ t ≤ 4 h(1)=(1,445.1) h(4)=(4,371.6)
= (371.6-445.1)/(4-1)
= -24.5m/s √
(2) 1 ≤ t ≤ 2 h(1)=(1,445.1) h(2)=(2,430.4)
= (430.4-445.1)/(2-1)
= -14.7m/s √
(3) 1 ≤ t ≤ 1.5 h(1)=(1,445.1) h(1.5)=(1.5, 438.98)
= (438.98-445.1)/(1.5-1)
= -12.25m/s √
(d) Use the secant method to approximate the instantaneous velocity at t=1 h(0.5) = (0.5, 448.78) h(1) = (1, 445.1)
= (445.1-448.78)/(1-0.5)
= -7.35m/s
(You must choose a point very close to (1, 445.1) in order to get an accurate answer. Try to choose values within 0.1 of the t-value given. Therefore, t = 0.9, or the point (0.9,446.031) is a good choice, and will result in an IROC of -9.31 m/s. This is very close to the true value of -9.8 m/s. As you can see, when t = 0.5, the answer is not as close to the true value.)
2. 6/7
The mass M in grams of undissolved sugar left in a teacup after t seconds is given by M =10.5-0.4t2
(a) When will all the sugar dissolve?
0=10.5-0.4t2
0.4t2=10.5 t2=26.25 t=5.12 sec
(b) Find the average rate of change in the interval 0 ≤ t ≤ 1
M(0) = (0, 10.5); M(1)= (1, 10.1)
= (10.1-10.5) / (1-0)
= -0.4 g/s
(Please make sure to include word answers or concluding statements, and to specify units. We are marking for communication and presentation, as well as for correct answers.)
(c) Draw on graphing paper a graph of M with respect to t and use the secant method to approximate the instantaneous rate of change at t = 2 seconds
M(1.5) = (1.5, 9.6); M(2)= (2, 8.9)
= (8.9-9.6)/(2-1.5) = -1.4 instantaneous rate of change at t = 2 seconds is -1.4
(Again, you must choose a value for t that is within ± 0.1
The height in metres of a ball dropped from the top of the CN Tower is given by h(t)= -4.9t2+450, where t is time elapsed in seconds.
(a) Draw the graph of h with respect to time
(b) Find the average velocity for the first 2 seconds after the ball was dropped h(0)=(0,450), h(2)=(2,430.4)
= (430.4-450)/(2-0)
= -9.8m/s √
(c) Find the average velocity for the following time intervals
(1) 1 ≤ t ≤ 4 h(1)=(1,445.1) h(4)=(4,371.6)
= (371.6-445.1)/(4-1)
= -24.5m/s √
(2) 1 ≤ t ≤ 2 h(1)=(1,445.1) h(2)=(2,430.4)
= (430.4-445.1)/(2-1)
= -14.7m/s √
(3) 1 ≤ t ≤ 1.5 h(1)=(1,445.1) h(1.5)=(1.5, 438.98)
= (438.98-445.1)/(1.5-1)
= -12.25m/s √
(d) Use the secant method to approximate the instantaneous velocity at t=1 h(0.5) = (0.5, 448.78) h(1) = (1, 445.1)
= (445.1-448.78)/(1-0.5)
= -7.35m/s
(You must choose a point very close to (1, 445.1) in order to get an accurate answer. Try to choose values within 0.1 of the t-value given. Therefore, t = 0.9, or the point (0.9,446.031) is a good choice, and will result in an IROC of -9.31 m/s. This is very close to the true value of -9.8 m/s. As you can see, when t = 0.5, the answer is not as close to the true value.)
2. 6/7
The mass M in grams of undissolved sugar left in a teacup after t seconds is given by M =10.5-0.4t2
(a) When will all the sugar dissolve?
0=10.5-0.4t2
0.4t2=10.5 t2=26.25 t=5.12 sec
(b) Find the average rate of change in the interval 0 ≤ t ≤ 1
M(0) = (0, 10.5); M(1)= (1, 10.1)
= (10.1-10.5) / (1-0)
= -0.4 g/s
(Please make sure to include word answers or concluding statements, and to specify units. We are marking for communication and presentation, as well as for correct answers.)
(c) Draw on graphing paper a graph of M with respect to t and use the secant method to approximate the instantaneous rate of change at t = 2 seconds
M(1.5) = (1.5, 9.6); M(2)= (2, 8.9)
= (8.9-9.6)/(2-1.5) = -1.4 instantaneous rate of change at t = 2 seconds is -1.4
(Again, you must choose a value for t that is within ± 0.1