Question

12. During the first 30seconds of its flight, a test rocket's height in metres above its launching point is given by h(t) = 452 – where t is the elapsed time in seconds. (a) Find an equation for the velocity of the rocket and use this to find how long it will take to reach 600m/s. (b) What height will the rocket be at this time? 13. Draw the graphs of the functions in question 12 in the same coordinate plane. You can use desmos.com. Explain the significance of the critical points of these functions and their connections. 14. An object's position is described by the following polynomial function for t=0 to t=10 s. y = - 15t + 54t Where y is in meters, t is in seconds, and positive is forward. Determine: a) the object's velocity as a function of time b) the object's acceleration as a function of time c) the times when the object is not moving d) the time when the object was moving backward the times when the object returned to its starting position f) Draw the graphs of the functions in this question in the same coordinate plane. You can use desmos.com. Explain the significance of the critical points of these functions and their connections. 15. Suppose the amount of water in a holding tank is given by the equation v(t) = 22 - 16t+35 a) Is the volume of water in the tank increasing or decreasing at t = 1 minute? b) Is the volume of water in the tank increasing or decreasing at t = 5 minutes? c) Is the volume of water in the tank changing faster at t = 1 or t = 5 minutes? d) When is the volume of water not changing? 16. Find the equation of the tangent to the curves: a) y = 3x2 +5 at x = 1 b) f(x) = x at x = 9

Answer 1

i b) at 12. h(t) = 45+2-73 h(t)- 900-3+2 osts 30 V(t)= 5')= 9ot- 3t2 V[+)= 600 3+2 – 900+600 = t2 - 30t+200=0 (t-10)(-20) t-10,20 (sec) =) t=10inct) = 3500m t=20 ; nH) = 1000 om. 13. Gitical point of hCH hilH co-) t=0,30. =) velocity =0 at critical point of ncol. critical point of nickl. nult)=0 -) 90-64 20-) t = 15 acceleration is o cat critical point of nicel.

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14) y + 3-15+2 +5 4+ (m) Os t = 10 is). a) VCH) = yllt) = 3+2 – 301754. b) alt) = y(t) = 67-30 c) UCH)=0 =) 3+2-30++54 =0 +2-105+18=0 te 101 v1oo-72 2 = 5 +17 t=s- (sec) E = sts (sec). d) V'LE) <o tt (5-57, 5+7) e) y = 0 Y=+3–15+2+54€ =+(12-157+54) =t(t-6)(1-9) + =o=) += 5, 6, 9. t=6,9

-60 150 40 -30 20 10 8 10 11 12. -10 -20