These questions are pertaining to College Algebra

Question

1 a) A boat travels upriver to a landing and then returns to its starting point. The trip upriver takes 1.2 hour, and the trip back takes .9 hour. If the average speed on the return trip is 5mph faster than on the trip upriver, what is the boat's speed upriver?

 

b) A projectile is fired straight up from ground level. After t seconds, its height above the ground is s feet, where

 

                                                                                          s = - 16t^2 + 220t

 

When will the projectile be at least 624ft above the ground?

Answer 1

a) The distance there is equal to the distance back so you can write:

 

(1.2 hr)x = (0.9 hr)*(5 mi/hr)

 

solve for x, the average speed upriver.

 

(1.2 hr)x = 4.5 mi

 

x = (4.5 mi) / (1.2 hr)

 

x = 3.75 mi/hr

 

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b) You are given s = the distance above ground, and the equation for s in terms of t seconds.

to find the time, t when the projectile is 624 ft above ground, you plug it in and solve for t.

 

624 = -16t² + 220t

 

0 = -16t² + 220t - 624

 

Use the Quadratic Eq.  t = [-b ± √(b² - 4ac)] / 2a

 

a = -16

b = 220

c = -624

 

t = [-220 + √((220)² - 4(-16)(-624))] / 2(-16)

 

t = [-220 + √(48400 - 39936)] / (-32)

 

t = [-220 + √(8464)] / (-32)

t = [-220 + 92] / (-32)

 

t = (-128) / (-32)

 

t = 4 seconds

 

t = [-220 - 92] / (-32)

 

t = (-312) / (-32)

 

t = 9.75 seconds

 

There will be two times because 624 ft is not the maximum height the projectile reaches. It reaches 624 ft at 4 seconds on the way up and 624 ft at 9.75 seconds on the way down.