Question

A car is crossing a road-turn with a speed of 10 metre/
second. If the coefficient of friction be 0.5, then the
radius of the turn of the car is (g = 10 m/s):
(a) 20 metre
(b) 10 metre
(c) 5 metre
(d) 2 metre​

Answer 1

Answer:

• Radius of turn of the Car = 20 metres

Explanation:

Given:

• Velocity of Car while crossing the road turn, v = 10 m/s
• Coefficient of friction, μ = 0.5
• Acceleration due to gravity to be taken, g = 10 m/s²

To find:

• Radius of the turn of the Car, R =?

Formula required:

• Relation between turning radius and speed of vehicle

        v = √( μ R g )

[ Where v is velocity, μ is coefficient of friction, R is turning radius and g is acceleration due to gravity ]

Calculation:

Using formula

→ v =√( μ R g )

→ v² = μ R g

→ ( 10 )² = ( 0.5 ) · R · ( 10 )

→ 100 = 5 · R

→ R = 100 / 5

→ R = 20 m

Therefore,

• Radius of the turn of car would be 20 metres.

Answer 2

$\underline{\underline{\sf{\color{orange}{GIVEN:-}}}}$

• A car is crossing a road-turn with a speed, v = 10 m/s
• The coefficient of friction, μ = 0.5
• Acceleration due to gravity, g = 10 m/s²

$\underline{\underline{\sf{\color{orange}{TO\ DETERMINE:-}}}}$

The radius of the turn(R) of the car.

$\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

We know,

$\boxed{\bf{v=\sqrt{\mu Rg} }}$

Solving further by substituting values :-

$\sf{\longmapsto 10=\sqrt{\mu Rg} }$

$\sf{\longmapsto 10=\sqrt{0.5 \times R \times 10} }$

$\sf{\longmapsto 10=\sqrt{5R} }$

$\sf{\longmapsto (10)^2=(\sqrt{5R})^2}$

$\sf{\longmapsto 100=5R}$

$\sf{\longmapsto R=\dfrac{100}{5}}$

$\sf{\longmapsto R=20\ m}$

∴ The radius of the turn of the car is 20 m.

The answer is (a) 20 meter.