$\begin{array}{1 1} 1.4 \times 10^2\;m \\ 2.4 \times 10^2\;m \\ 1.9 \times 10^3\;m \\7.4 \times 10^2\;m \end{array} $

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Answer : $1.4 \times 10^2\;m$

$\overrightarrow{V}_i=20.0\;m/s$

$\overrightarrow {a}=3.2 m/s^2$

$\Delta t= 5.0\;s$

$\Delta \overrightarrow{d}=\overrightarrow {v_i} \Delta t+ \large\frac{1}{2} $$\overrightarrow{a} \Delta t^2$

$\qquad= \bigg( 20.0 \large\frac{m}{g} \bigg)$$ (5.0\;g)+\large\frac{1}{2} $$\bigg(3.2 \large\frac{m}{g^2} \bigg) $$(5.0\;g )^2$

$\qquad= 100 m +40 m$

$ \Delta \overrightarrow{d} =1.4 \times 10^2m$

During the $5.0\;s$ time interval , the car is displaced $1.4 \times 10^{2} \;m$

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