91Ó°ÊÓ

$$\begin{gathered} \mathrm{s}=24\mathrm{m} \\ \mathrm{t}=2.00\mathrm{s} \\ {\mathrm{v}}_0=56\mathrm{km}/\mathrm{hr} \end{gathered}$$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using these equations, the magnitude of constant acceleration and the velocity when there is an impact can be calculated.

Formula:

In kinematic equation the displacement is given by

$∆x=v_{0}t+\frac{1}{2}a{t}^2$ (i)

The velocity is given by,

$v_f=v_0+a×t$ (ii)

Where $v_0$is the initial velocity

To find acceleration first convert 56.0 km /h into m/s

$$\begin{gathered} \frac{1\mathrm{km}}{\mathrm{h}}=\frac{1000}{3600}\mathrm{m}/\mathrm{s} \\ \frac{56\mathrm{km}}{\mathrm{h}}×\frac{1000}{3600}\frac{\mathrm{m}}{\mathrm{s}}×\frac{\mathrm{h}}{1\mathrm{km}}=15.55\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (i), the acceleration can be written as,

$$\begin{gathered} 24=15.55×2.00+(0.5×\mathrm{a}×2.{00}^2) \\ \mathrm{a}=-3.56\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Magnitude of acceleration $3.56\mathrm{m}/{\mathrm{s}}^2$.

Using equation (ii) the velocity will be,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}×\mathrm{t} \\ =15.55-(3.56×2) \\ =8.43\mathrm{m}/\mathrm{s} \end{gathered}$$

Final velocity of the car before the impact, ${\mathrm{v}}_{\mathrm{f}}=8.43\mathrm{m}/\mathrm{s}$.