Aballoon is rising vertically above a level, straight road at a constant rate of 2 ft divided by sec. just when the balloon is 70 ft above the ground, a bicycle moving at a constant rate of 19 ft divided by sec passes under it. how fast is the distance s (t )between the bicycle and balloon increasing 3 seconds later?

13 ft/s

Explanation:

Given

Balloon velocity is 2 ft/s upwards

Distance between balloon and cyclist is 70 ft

Velocity of cyclist is 19 ft/s

After 3 sec

cyclist traveled a distance of

Distance traveled by balloon in 3 s

net height of balloon from ground =6+70=76 ft[/tex]

at

distance between cyclist and balloon is

now suppose at any time t cyclist cover a distance of x m and balloon is at a height of h m

thus

differentiating w.r.t time

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.