# A sports car approaches a highway on-ramp at a velocity of $20.0\;m/s$ . If the car accelerates at a rate of $3.2 \;m/s^2$ for $5.0\;s$ , What is the displacement of the car ?

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

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 $1.4 \times 10^{2} \;m$