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YumYum247
 one year ago
someone please check my work......Thanks
YumYum247
 one year ago
someone please check my work......Thanks

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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2@nincompoop please cheque my work....:)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2does my work make any sense to you????????o_O

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2that's such a political response but i'll take that as a yes!!!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think nincompoop's profile picture matches your question and it's tilting it's head to read what you wrote :P

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2@IrishBoy123 please help

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2i;ve already done the question but the answer is different

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2what have i done wrong??????

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1well the car is deaccelerating so u need to take the acceleration as 2.3m/s^2 :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2ok let me try the question that way and i;ll get back to you

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.1try to think this way  if the car ws accelerating instead of deaccelerating then it wuld have certainly covered more distance. so if u try to calculate the distance covered by the accelerating car u wuld apply this formula S=ut +1/2 at^2 but u r applying the same formula for deaccelrating car so u r getting the same distance nd this is not possible so u can conclude that u made a mistake while taking the sign of acceleration :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2got it......i got the answer Thanks man :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2@Astrophysics LOL my question was pretty easy, maybe she has a sprained neck or something.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, yes, kinematics problems aren't too difficult, the problem most people have with this is, the terms, and knowing what they mean. it can get confusing I guess, when you're dealing with average velocity, instantaneous velocity, displacement, distance, deceleration, etc =P

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So for now you can think of deceleration as negative acceleration

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0deceleration =  a

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2got it...actually i've this this unit before but yah i guess i forgot about this particular thing LOL :D

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, yeah but if you want to be technical, deceleration really just means decrease in speed, I really don't like this term though xD.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.2me neither....!!! XD
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