Question #226623

A hovercraft stays off the ground by blowing air out underneath it. It can accelerate (forwards or backwards) at 0,6 m/s2 by blowing air with a steering fan (backwards or forwards). If the hovercraft is moving forwards at 12 m/s and the steering fan is suddenly turned to accelerate the hovercraft backwards, calculate:

a) the velocity of the hovercraft after 2 s

b) the time it takes for the hovercraft to reach zero velocity 

c) the velocity of the hovercraft after 29 s

d) the displacement of the craft when it has zero velocity

e) the displacement of the hovercraft when it is moving backwards at 16m/s

f) Clearly show what the value of h(max),t(total) and v

State whether the hovercraft is going backwards or forwards after each question’s answer.

(Make forward direction positive and backwards direction negative)




1
Expert's answer
2021-08-17T16:54:02-0400


Part (a)

we have to find velocity after 2sec is:

v=u+atv=120.62=10.8m/sv=u+at\\ v=12-0.6*2=10.8m/s

(here a=-0.6m/s^2 because the hovercraft suddenly start moving backward)

Part (b)

we have to find time taken by hovercraft when its velocity become zero:

using same formula:

v=0,u=12m/s,a=0.6m/s2v=0,u=12m/s,a=-0.6m/s^2

so,t=20secso, t=20sec


Part (c)

velocity after 29sec is:

v=12(0.629)=2.6m/sv=12-(0.6*29)=2.6m/s


Part (d)

displacement when velocity becomes zero:

v2=u2+2ass=(u2)2a=(1212)(2.6)=120mv^2=u^2+2as\\ s=\frac{(u^2)}{2a}=\frac{(12*12)}{(2*.6)}=120m


Part (e)

displacement when it is moving backward at velocity 16m/s

sos=(v2u2)2as=(16161212)(20.6)=93.3mso s=\frac{(v^2-u^2)}{2a}\\ s=\frac{(16*16-12*12)}{(2*0.6)}=93.3m


Part(f)

v=u+atv=u+at (1st eqaution of motion)

v=0,u=12m/sanda=0.6v=0,u=12m/s and a=-0.6

therefore,t=20sect=20sec (time of going upward)

hence, total time of flight is 220=40sec2*20=40 sec

hmax=(u2)2a=(1212)(2.6)=120mh_{max}=\frac{(u^2)}{2a}=\frac{(12*12)}{(2*.6)}=120m

v=st=12040=3m/sv= \frac{s}{t}=\frac{120}{40}= 3 m/s


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