91Ó°ÊÓ

We know that the radiation of curvature of the road is $r=200\mathrm{m}$, speed of the car is $v=25\mathrm{m}/\mathrm{s}$, and the coefficient of kinetic friction is $\mathrm{μ}=1.0$.

We have to calculate the acceleration of the car after applying the brake.

Consider the mass of the car is $m$. Hence, the weight of the car is $w=mg$, where $g=9.8\mathrm{m}/{\mathrm{s}}^2$is the acceleration due to gravity.

localid="1648323021846" $N=mg$

The force due to friction is

$F_A=\mathrm{μ}N=\mathrm{μ}mg$

Therefore, the acceleration is

$a_A=\frac{\mathrm{μ}mg}{m}=\mathrm{μ}g$

The acceleration of the car after applying the brake is:

$a_A=(1.0)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)=9.8\mathrm{m}/{\mathrm{s}}^2$.

Now car B is at the bottom of the hill. Therefore, the centrifugal force acting on the car will be downwards and the total vertical force acting on the car will be

$N-mg=\frac{m{v}^2}{r}$

$N=m\left(g+\frac{v^2}{r}\right)$

Hence, the acceleration is given by

$a_B=g+\frac{v^2}{r}$

Now substituting the values we have

$a_B=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+\frac{(25\mathrm{m}/\mathrm{s}{)}^2}{(200\mathrm{m})}$

localid="1648323043916" $=12.925m/s^2$

Now for the car C is at the top of the hill. Therefore, the centrifugal force will act upward and the total vertical force on the car is:

$mg- N=\frac{m{v}^2}{r}$

Now, similarly the acceleration of the car will be

$a_C=g-\frac{v^2}{r}$

substituting the values we have

$a_B=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\frac{(25\mathrm{m}/\mathrm{s}{)}^2}{(200\mathrm{m})}$

$=6.7\mathrm{m}/{\mathrm{s}}^2$

The magnitude of tangential acceleration of car A is $9.8\mathrm{m}/{\mathrm{s}}^2$, car B is localid="1648323058991" $12.925\mathrm{m}/{\mathrm{s}}^2$and car C is $6.7\mathrm{m}/{\mathrm{s}}^2$.