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)




Expert's answer


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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