Question

A blimp is ascending in the air at a speed of 4.28 m/s when the pilot turns off the engine. The blimp immediately begins to experience constant acceleration, such that in its ascent, stops for an instant, and begins to sink. The blimp is at its highest point 10.2 s after the engine is turned off. (It is a blimp full of helium, so even with engines off it falls gently, it does not drop like a rock) A. How far has the blimp ascended, at the moment when it makes a momentary stop?

B. How long will it take to get back to the height at which the engine was turned off?
C. What will its speed be when it passes through that original height again?
D. Graph position vs time (y vs t), velocity vs time, and acceleration vs time for the entire up-and-down trip. (Include axis labels, marked positions and times, etc.)

Answer 1

a) 21.8 mts

b) 10.2 seconds

c) 4.28 m/s

Step-by-step explanation:

Because the blimp is filled with helium, the acceleration won't be the gravity. We have to calculate the new acceleration:

`a=(Vf-Vo)/(t)\\\\a=(0-4.28)/(`10.2)\\\\a=0.420 m/s^2`

in order to obtain the height we have to use the formulas of accelerated motion problems:

`X=Vo*t+(1)/(2)*a*t^2\\\\X=4.28*(10.2)+(1)/(2)*(-0.420)*(10.2)^2\\X=21.8mts`

we can calculate the time with the same formula:

`-21.8=0*t+(1)/(2)*(-0.420)*t^2\\solving\\t=10.2 seconds`

the velocity at the same height is given by:

`Vf^2=Vo^2+2*a*x\\Vf=√(2*(-0.420)*(-21.8)) \\Vf=-4.28m/s`

the speed would be 4.28m/s because is a scalar value.