A balloon is rising vertically at 1ft/sec, when it is 65ft above ground a bike goin 17ft/sec passed under it?

Question

How fast is the distance between the bicycle and the ballon INCREASING 3 seconds later?

Using Calculus...antidifferentiation or some other calc formula...How fast is the distance between them "INCREASING" 3 seconds later

Answer 1

quickly

Answer 2

a^2 + b^2 = c^2 is the key.

The balloon will be 68ft in the air and the bike will be 51ft away from directly below the ballon. The straight line distance from the balloon will be 85ft. This is based on the Pythagorean theory shown above.

So, the rate of change will be:
(51/85) *17 for the rate the bike is moving (51/85 is the COS of the angle of displacement between the bike and the balloon)
and
(68/85) * 1 for the rate the balloon is moving.

Add them together and find that the distance is moving at:
10.2ft/sec + 0.8ft/sec for 11ft/sec

Answer 3

The best way to solve this would be to first draw a picture/diagram of what is happening initially and then draw another diagram where the object will be 3 seconds later, write down what numbers you are giving like velocity of the ballon= 1ft/sec, and t0=0, t1=3, velocity of the bike 17ft/sec, and the initial location of each object and analyze it from there using the equations from your text or classroom.

Answer 4

85 feet.

65 seconds later, the balloon is 65 feet in the air and then three seconds after the bike is directly under it, it was 68 feet in the air. the bike was 51 feet past the point at which it was directly under the balloon. so you draw a triangle and use the formula a²+b²=c²(pythagorean's theorem). so this means that 68²+51²=c². 4624+2601=c². 7225=c². then you do the square root on both sides to cancel out the square and get the square root of 7225 which is 85. so three seconds after the bike passed under the balloon, the balloon and the bike were 85 feet apart.