during a pond hockey game, a punk accelerates from rest at 5.0m/s/s over a distance of 80.0cm. the punk then slides with a constant speed for 4.0s until it reaches a rough section which causes it to stop in 2.5 sec.

what is the speed of the object when it reaches the rough section.

Helllo........ anybody to help. thanks

Question

during a pond hockey game, a punk accelerates from rest at 5.0m/s/s  over a distance of 80.0cm. the punk then slides with a constant speed for 4.0s until it reaches a rough section which causes it to stop in 2.5 sec.

what is the speed of the object when it reaches the rough section.

Helllo........ anybody to help. thanks

anonymous · Answer

Vf^2 = Vi^2 + 2ad so Vf = (2x5x0.80)^0.5 = 2.83 m/s into the rough.

khally92 · Answer

how i do not understand

ghazi · Answer

for the distance of 80 cms you need to use $$V^2=U^2+2as=2*5*0.8=8$$ so V that is the velocity of that punk when it reaches to the beginning of the sliding motion will be $$V= 2 \sqrt{2}m/s$$ after that punk slides with this velocity because it is the velocity with which he touches the rough surface so its your answer and refer the diagram |dw:1356304048907:dw| up to point B velocity remains constant and after that it starts decreasing between A and B punk is sliding with constant speed , so thats your answer

khally92 · Answer

Hi Thanks i waited for you to come online, this is the B part of the question its like i understand the question now, 

At what rate does the object slow down once it reaches the rough section? i guess that is acceleration huhh? well i got -0.71. am i right.

khally92 · Answer

i used V=U+at

ghazi · Answer

thats correct

khally92 · Answer

What total distance does the object slide through out its entire trip?

ghazi · Answer

s= ut + 1/2 a t^2 and u =v

khally92 · Answer

which U=V

ghazi · Answer

when the sliding starts that velocity is V and used has U here

khally92 · Answer

the question asked fot tatal distance is it not suppose to be  0.8+ sliding+ into the rough.

khally92 · Answer

i dont get this

khally92 · Answer

what is the distance for constant velocity? distance = velocity*time. what is the constant velocity?

ghazi · Answer

find distance up to A first, then distance up to B , for distance A to B you have to use S=velocity* time , because from A to B velocity is constant, and after that from point B your initial velocity = the ending velocity of sliding motion and final would be zero because it stops

khally92 · Answer

is it 2.83/4 =0.07075

ghazi · Answer

i think so :) i havent done calculation , i just told you whole process :)

ghazi · Answer

distance A to B = 2.83*4=11.31 and after that next distance = ut + 1/2 a t^2

khally92 · Answer

what is a for rough surface -o.70

khally92 · Answer

total distance o.8+11.31+4.87=16.98

ghazi · Answer

that is right i think :)