Rohangrr
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



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baddinlol
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Does it start from rest?

Rohangrr
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dw:1342006345931:dw

Rohangrr
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thanks

baddinlol
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Wait

baddinlol
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Its wrong

baddinlol
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Use the formula v = ut + 0.5at^2

Shane_B
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Assuming a start from rest: \[d=Vi(t)+\frac{1}{2}at^2=0m/s^2+\frac{1}{2}(4m/s^2)(10s)^2=200m\]

baddinlol
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Oops sorry

Rohangrr
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lol yes firstly u found the mistake

baddinlol
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x = ut + 0.5at^2

Rohangrr
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@Shane_B can u make it more clearer

Shane_B
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Or, you can use: \[d=\frac{V_i+V_f}{2}t=\frac{0m/s+40m/s}{2}(10s)=200m\]

Shane_B
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Which part isn't clear?

Rohangrr
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well plzzz make it clear from the beggining

baddinlol
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Well he used this formula x = ut + 0.5at^2

baddinlol
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u = 0m/s , becausehe assumed the car started from rest

baddinlol
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t = 10s

baddinlol
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a = 4m/s^2

baddinlol
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Then he plugged in the numbers x = 0.5*4* 100 = 200m

Shane_B
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@baddinlol: thanks...working with someone else at the moment :)

Rohangrr
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@Shane_B ITS \[\huge 4 \quad m/s^2 \]

Rohangrr
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not 40 m/s2

baddinlol
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no no

baddinlol
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The final velocity is 40m/s

Shane_B
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40m/s is the final velocity...

Rohangrr
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okay lol