91Ó°ÊÓ

The initial speed of the truck: $25\mathrm{m}/\mathrm{s}$.

Distance of the hole from moving truck: $55\mathrm{m}$.

Reaction time : $0\mathrm{s}$.

$v^2=u^2+2as$.

In order to determine whether we can stop the truck before it falls into the hole, we have to use the equation of the kinematics $v^2=u^2+2as$

The truck comes to rest, so its final speed be$0\mathrm{m}/\mathrm{s}$.

Now substituting the various values to the equation,

We get,

$\begin{array}{r}{v}^2=u^2+2as\\ 0=(25{)}^2−2(a)55\end{array}$

Here, the$a^2$is the rate of retardation.

Now solving for the value of $a$,

We get

$\begin{array}{r}0=(25{)}^2−2(a)55\\ a=\frac{625}{110}=5.68\mathrm{m}/{\mathrm{s}}^2\end{array}$

So, for the truck to stop before it falls into the hole, the minimum retardation or de-acceleration required is$5.68\mathrm{m}/{\mathrm{s}}^2$.

The initial speed of the truck: $25\mathrm{m}/\mathrm{s}.$

Coefficient of static friction ${\mathrm{μ}}_s:0.60$.

Coefficient of kinetic friction ${\mathrm{μ}}_k:0.30$.

$\begin{array}{rl}& F={\mathrm{μ}}_{s}N\\ & v^2=u^2+2as\end{array}$.

The static friction force can be given as $F={\mathrm{μ}}_{s}N$.

Further, we know that the force is given as $F=ma$.

By equating both the equations as,

$\begin{array}{c}ma={\mathrm{μ}}_{s}N\\ \text{or,}ma={\mathrm{μ}}_{x}mg\\ \text{or,}a={\mathrm{μ}}_{s}g\end{array}$

This value of $a^2$will give the rate of retardation for which the truck stop without the antiques sliding and being damaged.$\begin{array}{rl}& \text{So,}a={\mathrm{μ}}_{s}g\\ & a=0.60×9.8=5.88\mathrm{m}/{\mathrm{s}}^2\end{array}$

If we de-accelerate the truck with $5.88/{\mathrm{s}}^2$, the truck can stop without the antiques sliding and being damaged.