A particle starts from rest and moves with constant acceleration $a$. Which expression gives its displacement from initial position $x_0$ after time $t$?
A$x=x_0+at^2$
B$x=x_0+\dfrac{1}{2}at^2$
C$x=x_0+vt$
D$x=x_0+\dfrac{1}{2}vt^2$
Answer & Solution
Correct answer: B. $x=x_0+\dfrac{1}{2}at^2$
Using $x=x_0+v_0t+\dfrac{1}{2}at^2$ and $v_0=0$ for starting from rest, we get $x=x_0+\dfrac{1}{2}at^2$. Option A misses the factor $\tfrac12$.
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