Question

A car is travelling at 57.6 km/hr on an unbanked (horizontal) circular road of radius om. If the
160/3
coefficient of friction between the road and the car is 0.8, then the maximum tangential deceleration
that driver of car can achieve by applying the brakes at this moment is : (g = 10 m/s-)​

Answer 1

Answer:6.4 metre per second square

Explanation:

This is a typical problem of centripetal force with car on an unbanked road. The forces acting on the car are

Weight of the car mg acting downward.

Reaction forces in upwards direction on the two tires R1 and R2

Horizontal frictional forces that opposes centrifugal forces inwards direction.

Now total vertical reaction can balance the weight

R1+R2 =mg

The total friction Will be the centripetal forces.

F1+F2=mv²/r

As static friction is self adjustable

F1+F2<= Fs = u_s(R1+R2) = u_s mg

Let u_s be coefficient of static friction.

Hence we have

mv²/r<= u_s mg

v<= √ u_s great

If v excceds this value it will skid out. Hence it is max limit.

Now for inculcating deceleration.

umg = m√(v/r)²+ a²

Putting the values of r= 160/3 v=57.6*5/18= 16ms-1 u= 0.8 g=10

We have

a = 6.4 m/s²

Answer 2

Answer:

6.4 meter per second squared

Explanation:

 According to this question

We have been given that

This is a typical problem of centripetal force with a car on an unbanked circular road. The forces acting on the car are

Weight of the car mg acting downward.

Reaction forces in upwards direction on the two tires $R_1$ and $R_2$

Horizontal frictional forces that oppose centrifugal forces inwards direction.

Now total vertical reaction can balance the weight

$R_1+R_2$ = mg

The total friction will be the centripetal forces.

$F_1 + F_2$ = $m\times \frac{v^2}{r}$

As static friction is self adjustable

$F_1 + F_2 \leq F_s$ = $u_s(R_1+R_2)$ = $u_s mg$

Let $u_s$ be the coefficient of static friction.

we have

$m\frac{v²}{r} \leq u_s mg$

$v \leq √ u_s$ greater

If v exceeds this value it will skid out. Hence it is maxed limit.

Now for inculcating deceleration.

$u\times m\times g$ = $m\times \sqrt(\frac{v}{r})^2 + a^2$

Putting the values of r= $\frac{160}{3}$ = 53.34 v=$57.6\times \frac{5}{18}$= 16ms-1 u= 0.8 g=10

$0.8\times m\times 10$ = $m\times \sqrt[\frac{16}{53.34}]^2+ a^2$

$a^2$ =  40.96

a = 6.4 m