4. A model rocket moves in the zy plane (the positive y-axis points upward and p

Question

4. A model rocket moves in the zy plane (the positive y-axis points upward and perpendicular to the ground). The rocket's velocity has components: The constants are: A 1.0 m/s2, B 9.0 m/s3, C 2.0 m/s, D 4.0 m/s2 and E 12.0 m/s At t 0 the rocket is at the origin (a) Calculate the components of the acceleration vector of the rocket as functions of time lie. ar (t) and av(t)]. (Present details of your calculation) (8 pts.) (b) Calculate the components of the position vector of the rocket as functions of time lie. r() and (t)]. (Present details of your calculation) (8 pts.)

Explanation / Answer

vy=A*t+B*t^2+C*t^3

=t+9*t^2-2*t^3

then y=integration of vy*dt

=0.5*t^2+3*t^3-0.5*t^4+c

where c is constant of integration

now at t=0, y=0

hence c=0

y=0.5*t^2+3*t^3-0.5*t^4

vx=4*t+12*t^2

then x=integration of vx*dt

=2*t^2+4*t^3

part C:

at maximum height,

dy/dt=0 and d^2y/dt^2 <0

dy/dt=t+9*t^2-2*t^3

equating dy/dt=0

==>t*(1+9*t-2*t^2)=0

==>t=0 or t=4.6085 or t=-0.1085

0 and -0.1085 are not possible.

hence at t=4.6085 seconds, it reaches maximum height

part D:
when the rocket hits the ground, y=0

==>0.5*t^2+3*t^3-0.5*t^4=0

==>t^2*(0.5+3*t-0.5*t^2)=0

==>t=0 or t=-0.1623 or t=6.1623

so correct solution is t=6.1623 seconds.

then horizontal distance=2*6.1623^2+4*6.1623^3=1011.975 m

part E:

ax=dvx/dt=D+2*E*t=4+24*t

at t=6.1623, ax=4+24*6.1623=151.8952 m/s^2

ay=dvy/dt=1+18*t-6*t^2

at t=6.1623 seconds, ay=1+18*6.1623-6*6.1623^2=-115.922 m/s^2

magnitude of acceleration=sqrt(ax^2+ay^2)

=sqrt(151.8952^2+115.922^2)=191.076 m/s^2

angle with +ve x axis=arctan(ay/ax)=arctan(-115.922/151.8952)=-37.35 degrees

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