Q. 16

# A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30^{o} (figure 6-E5). The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes; the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s^{2}.

Answer :

The maximum force of friction is developed between the surface of road and the car’s tyres when hard breaks are applied.

So, maximum frictional force = μ*R*

From the free body diagram,

*R* − *mg* cos θ = 0

⇒ *R* = *mg* cos θ (i)

and*μ R + ma − mg sin θ = 0 (ii)*⇒ μ

*mg*cos θ +

*ma*−

*mg*sin θ = 0

⇒ μg cos θ + a -10(12) =0

⇒ a=5- {1-(2)} ×10(/2)

⇒ a = -2.5 m/s^{2}

*When brakes are applied, car will deaccelerate by* 2.5 m/s^{2}

*Distance s*= 12.8 m*initial velocity u* = 6 m/s

∴ Velocity at the end of incline

ν=

v=10 m/s=36 km/h

Hence, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h

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PREVIOUSThe friction coefficient between an athelete’s shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop?NEXTA car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 s.

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