anonymous
  • anonymous
okay @lgbasallote @Shane_B @baddinlol A racing car has a uniform acc.. of \[ 4 \quad m/s^2\]. What's the distance she will cover 10s later
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Does it start from rest?
anonymous
  • anonymous
|dw:1342006345931:dw|
anonymous
  • anonymous
thanks

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anonymous
  • anonymous
Wait
anonymous
  • anonymous
Its wrong
anonymous
  • anonymous
Use the formula v = ut + 0.5at^2
Shane_B
  • Shane_B
Assuming a start from rest: \[d=Vi(t)+\frac{1}{2}at^2=0m/s^2+\frac{1}{2}(4m/s^2)(10s)^2=200m\]
anonymous
  • anonymous
Oops sorry
anonymous
  • anonymous
lol yes firstly u found the mistake
anonymous
  • anonymous
x = ut + 0.5at^2
anonymous
  • anonymous
@Shane_B can u make it more clearer
Shane_B
  • Shane_B
Or, you can use: \[d=\frac{V_i+V_f}{2}t=\frac{0m/s+40m/s}{2}(10s)=200m\]
Shane_B
  • Shane_B
Which part isn't clear?
anonymous
  • anonymous
well plzzz make it clear from the beggining
anonymous
  • anonymous
Well he used this formula x = ut + 0.5at^2
anonymous
  • anonymous
u = 0m/s , becausehe assumed the car started from rest
anonymous
  • anonymous
t = 10s
anonymous
  • anonymous
a = 4m/s^2
anonymous
  • anonymous
Then he plugged in the numbers x = 0.5*4* 100 = 200m
Shane_B
  • Shane_B
@baddinlol: thanks...working with someone else at the moment :)
anonymous
  • anonymous
@Shane_B ITS \[\huge 4 \quad m/s^2 \]
anonymous
  • anonymous
not 40 m/s2
anonymous
  • anonymous
no no
anonymous
  • anonymous
The final velocity is 40m/s
Shane_B
  • Shane_B
40m/s is the final velocity...
anonymous
  • anonymous
okay lol

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