A child rides down the road on his bike at a speed 4.3 m/s with a balloon tied t

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 1/10th 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.]

Explanation / Answer

given:

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

then

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

y-component, ay = 4.71 - 0*t

Part a):

Let x-component of speed be vx.

then ax = dvx/dt

vx = integration[ax*dt]

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

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

where C is a constant

at t = 0, vx = 4.3 m/s

C = 0

let at t = T, vx = (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 vy.

then vy = integration[ay*dt]

vy = integration[4.71 dt]

vy = 4.71 t + C

where C is a constant

at t = 0, vy = 0

C = 0

then

vy = 4.71 t

if height at time t is y,

then dy/dt = vy

y = integration[vy*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

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