For a model rocket rising vertically, the height at time t is given by S = 120t

Question

For a model rocket rising vertically, the height at time t is given by S = 120t - 16t^2. Find the greatest height the rocket attains. A. 120 B. 360 C. 225 D. 335 If y = cos (x), then dy/dx = (A) sin (x) (B) 1/sec (x) (C) -tan (x) cos (x) (D) sec (x) sin (x) Find equation of tangent to the curve y = x^2 -3x +4 at x = 2 A. y - x - 2 = 0 B. y - 2x + 1 = 0 C. y = x D. y = -x Problem 15-16 are based on the following statement. The position of an object as a function of time is described by: x = 4t^3 + 2t^2 -t + 3. What is the acceleration of the object at t = 2? A. 26 B. 41 C. 52 D. 28 What is the total distance traveled from t = -2 to t = 2? A. 61 B. 67 C. 118 D. -421/27 A panicle travelled in a straight line in such a way that its distance S from a given that line after time t was S = 20t^3 - t^4. The rate of change of acceleration at time A. 72 B. 144 C. 208 D. 192

Explanation / Answer

12)

given S=120t -16t2

dS/dt=120-32t ,d2S/dt2=-32

at greatest height dS/dt=0,d2S/dt2<0

120-32t=0

=>t=120/32

=>t=3.75

greatest height rocket attains=S(3.75)

greatest height rocket attains=(120*3.75) -(16*3.752)

greatest height rocket attains=225

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13)y=cos(x)

dy/dx= -sin(x)

dy/dx= -(sin(x)/cos(x))cos(x)

dy/dx= -tan(x)cos(x)

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14)

y=x2-3x+4,x=2

y=22-(3*2)+4

y=4-6+4

y=2

point on curve is (2,2)

dy/dx=2x-3

at x=2

slope=(2*2)-3

slope=1

equation of tangent with slope 1 at point (2,2) is

y-2=1(x-2)

y=x

equation of tangent to the curve is y=x

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