Question

A child rides down the road on his bike at a speed 4.3 m/s with a balloon tied to its handle bars. At a time t = 0 s, the balloon comes untied and floats into the air. It accelerates with the vector:

a(t) = ( -0.31 m/s² e^{(-0.072093 s-1 t )} , 4.71 m/s² - 0 m/s³ t )

[Note: The acceleration coefficient in the x-direction should be v₀? (= -0.3099 m/s²) where v₀ is the initial speed and ? is the exponent; the acceleration and initial speed are related quantities.]

(a) What is the time it takes for the balloon to reach 1/10^th its original horizontal speed?


(b) How high does it rise in that time?  

[Also note: the parameters are too free in this problem, so about 1/8th of you will get a negative number for part (b); I don't want to change it since some people have already tried the problem.]

Answer 1

given:

a(t) = (-0.31*e(-0.072093*t), 4.71 - 0*t)

then

x-component, a_x = -0.31*e(-0.072093*t)

y-component, ay = 4.71 - 0*t

Part a):

Let x-component of speed be v_x.

then a_x = dv_x/dt

v_x = integration[a_x*dt]

= integration of -0.31*e(-0.072093*t)

= 4.3*e(-0.072093*t) + C

where C is a constant

at t = 0, v_x = 4.3 m/s

C = 0

let at t = T, v_x = (1/10)^th of original speed

4.3*e(-0.072093*T) = 0.1*4.3

T = 31.94 seconds

hence at 31.94 seconds, horizontal speed will reach 1/10th of its original horizontal speed.

Part b):

let y-component of speed be v_y.

then v_y = integration[a_y*dt]

v_y = integration[4.71 dt]

v_y = 4.71 t + C

where C is a constant

at t = 0, v_y = 0

C = 0

then

v_y = 4.71 t

if height at time t is y,

then dy/dt = v_y

y = integration[v_y*dt]

y = integration[4.71t dt]

y = 2.355*t² + C

at t = 0, y=0

C = 0

hence

y = 2.355*t²

at t = 31.94 seconds, height, y = 2402.49 m

Hence,

the baloon rises to a height of 2402.49 m