A car travelling 56.0 km/ his 24.0 mfrom the barrier when the driver slams on the brakes. The car hits the barrier 2.00 s later. (a)What is the magnitude of the car’s constant acceleration before impact? (b)How fast is the car traveling at impact?

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

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,

$v_f=v_0+a\times t$ (ii)

Where $v_0$is the initial velocity

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

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

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

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

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

Using equation (ii) the velocity will be,

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

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