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A particle starts from origin at $t=0$ with a velocity $5.0 \hat{ i }\; m / s$ and moves in $x-y$ plane under action of a force which produces a constant acceleration of $(3.0 \hat{ i }+2.0 \hat{ j })\; m / s ^{2} .$
$(a)$ What is the $y$ -coordinate of the particle at the instant its $x$ -coordinate is $84 \;m$ ?
$(b)$ What is the speed of the particle at this time?
Solution
Answer From Eq. $(4.34a)$ for $r _{0}=0$, the position of the particle is given by
$r (t)= v _{ o } t+\frac{1}{2} a t^{2}$
$=5.0 \hat{ i } t+(1 / 2)(3.0 \hat{ i }+2.0 \hat{ j }) t^{2}$
$=\left(5.0 t+1.5 t^{2}\right) \hat{ i }+1.0 t^{2} \hat{ j }$
$\text {Therefore, } \quad x(t)=5.0 t+1.5 t^{2}$
$y(t)=+1.0 t^{2}$
Given $x(t)=84 m , t=?$
$5.0 t+1.5 t^{2}=84 \Rightarrow t=6 s$
At $t=6 s , y=1.0(6)^{2}=36.0 m$
Now, the velocity $v =\frac{ d r }{ d t}=(5.0+3.0 t) \hat{ i }+2.0 t \hat{ j }$
At $t=6 s , \quad v =23.0 \hat{ i }+12.0 \hat{ j }$
speed $=| v |=\sqrt{23^{2}+12^{2}} \cong 26 m s ^{-1}$
