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A particle moves in a straight line. Its position ( in m) as function of time is given by `x = (at^2 + b)` What is the average velocity in time interval ` t = 3s to t = 5s in ms^(-1)`. (where a and b are constants and a `= 1ms^(-2), b = 1m`). |
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Answer» Correct Answer - 8 `V_(av) = (x_(f)- x_(i))/(t_(f)-t_(i)) = ((1 xx 5^2 +1)-(1xx3^2+1))/(5-3) = 16/2 = 8 ms^(-1)`. |
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