Question

When a derivative is taken of a VECTOR quantity, then the derivative of each COMPONENT is taken with respect to the same PARAMETER, i.e, d(x,y)/dt = (dx/dt,dy/dt). When objects travel on CURVED PATHS, and alternate RECTANGULAR description of the motion is the coordinates RELATIVE to the path. This has the advantage of immediately answering questions that are important to us: the TANGENTIAL component speeds up and slows down the object and the NORMAL component turns it.

1. A child rides down the road on his bike at a speed 4.6 m/s with a balloon tied to its handlebars. At a time t = 0 s, the balloon comes untied and floats into the air. It accelerates with the vector:
a(t) = ( -0.18 m/s² e^{(-0.0391304 s-1 t )} , 3.83 m/s² - 0.803 m/s³ t )

[Note: The acceleration coefficient in the x-direction should be v₀? (= -0.1799 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)=(-0.18*exp(-0.0391304*t),3.83-0.803*t)

then x component=ax=-0.18*e^(-0.0391304*t)

y component=ay=3.83-0.803*t

part a:

x component of speed let be vx.

then ax=dvx/dt

==>vx=integration of ax*dt

=integration of -0.18*e^(-0.0391304*t)

=4.6*e^(-0.0391304*t)+c

where c is a constant

at t=0,vx=4.6 m/s

==>c=0

let at t=T, vx=1/10 th of original speed

==>4.6*e^(-0.0391304*T)=0.1*4.6

==>T=58.844 seconds

hence at 58.844 seconds, horizontal speed will reach 1/10th of its original horizontal speed.

part b:

let y component of speed be vy.

then vy=integration of ay*dt

=3.83*t-0.4015*t^2+c

where c is a constant

at t=0, vy=0

==>c=0

then vy=3.83*t-0.4015*t^2

if height at time t is y,

then dy/dt=vy

==>y=integration of vy*dt

=1.915*t^2-0.13383*t^3+c

at t=0.y=0==>c=0

hence y=1.915*t^2-0.13383*t^3

at t=58.844 seconds, height=-2.0638*10^4 m