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A car skids a distance of 54m on a dry road. What was the velocity of the car when the skid began if the coefficient of friction between the tires and the road is 0.46?
A) $16\dfrac{m}{s}$.
B) $22\dfrac{m}{s}$.
C) $28\dfrac{m}{s}$.
D) $34\dfrac{m}{s}$.
E) Velocity cannot be determined with the information given.

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Last updated date: 05th Mar 2024
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IVSAT 2024
Answer
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Hint: The friction is the opposing force which acts on the body in the backwards direction. The body which is skidding is also facing a force of friction in the direction of motion. The final velocity of the car will be zero as the force of friction is continuously working on the car in the opposite direction.

Formula used:
The force of friction is given by,
$f = \mu mg$
Where force of friction is f the coefficient of friction is $\mu $ the mass of the car is m and the acceleration due to gravity is g.
The formula third relation of Newton’s law of motion is given by,
${v^2} - {u^2} = 2as$
Where final velocity is v, the initial velocity is u the acceleration is a and the distance is s.
The force on the car is given by,
$F = m \cdot a$
Where force on the body is equal to F, the mass of the body is m and the acceleration is a.

Complete step by step solution:
It is given in the problem that a car skids a distance of 54m on a dry road the coefficient of friction between the road and the tire of the car is 0.46 then we need to find the initial velocity of the car.
First of all let us find the acceleration of the car.
The force of friction is given by,
$ \Rightarrow f = \mu mg$
Where force of friction is f the coefficient of friction is $\mu $ the mass of the car is m and the acceleration due to gravity is g.
The acceleration of the car is equal to,
$ \Rightarrow a = \dfrac{f}{m} = - \dfrac{{\mu mg}}{m}$
$ \Rightarrow a = - \mu \cdot g$
$ \Rightarrow a = - \left( {0 \cdot 46} \right) \times \left( {9 \cdot 81} \right)$
$ \Rightarrow a = - 4 \cdot 5\dfrac{m}{{{s^2}}}$………eq. (1)
Now let us calculate the initial velocity of the car.
The formula third relation of Newton’s law of motion is given by,
$ \Rightarrow {v^2} - {u^2} = 2as$
Where final velocity is v, the initial velocity is u the acceleration is a and the distance is s.
The final velocity of the car is zero the acceleration of the car is $a = - 4 \cdot 5\dfrac{m}{{{s^2}}}$ the distance is equal to 54m.
$ \Rightarrow {v^2} - {u^2} = 2as$
$ \Rightarrow - {u^2} = 2 \times \left( { - 4 \cdot 5} \right) \times \left( {54} \right)$
\[ \Rightarrow {u^2} = 2 \times \left( {4 \cdot 5} \right) \times \left( {54} \right)\]
\[ \Rightarrow {u^2} = 486\]
\[ \Rightarrow u = \sqrt {486} \]
\[ \Rightarrow u = 22\dfrac{m}{s}\]

The initial velocity of the car is equal to \[u = 22\dfrac{m}{s}\]. The correct option for this problem is option (B).

Note: The acceleration acting on the car has sign negative because the force of friction is acting on the car and in the opposite direction of motion and therefore the acceleration of the car is in a negative direction. The kinetic energy of the car is used in the work done against the friction force.