In the case of constant frictional force when brakes are applied, how is the stopping distance affected?

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In the case of constant frictional force when brakes are applied, how is the stopping distance affected?

When examining the relationship between stopping distance and speed under the influence of a constant frictional force, the correct interpretation is that the stopping distance increases with the square of the speed. This can be understood through the concepts of work and energy. When brakes are applied, the vehicle’s kinetic energy must be dissipated by the work done against friction. The kinetic energy of a moving object is given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its speed. When the brakes are applied, friction does work to bring the vehicle to a stop, and the work done by friction can be represented as \( W = F_f \cdot d \), where \( F_f \) is the frictional force and \( d \) is the stopping distance. Since the work done by the frictional force must equal the loss of kinetic energy, we set these equal: \[ F_f \cdot d = \frac{1}{2}mv^2 \] From this relationship, we can solve for stopping distance \( d \): \[ d = \frac{\frac{1}{2}mv^2}{F_f} \

When examining the relationship between stopping distance and speed under the influence of a constant frictional force, the correct interpretation is that the stopping distance increases with the square of the speed. This can be understood through the concepts of work and energy.

When brakes are applied, the vehicle’s kinetic energy must be dissipated by the work done against friction. The kinetic energy of a moving object is given by the equation ( KE = \frac{1}{2}mv^2 ), where ( m ) is the mass of the object and ( v ) is its speed. When the brakes are applied, friction does work to bring the vehicle to a stop, and the work done by friction can be represented as ( W = F_f \cdot d ), where ( F_f ) is the frictional force and ( d ) is the stopping distance.

Since the work done by the frictional force must equal the loss of kinetic energy, we set these equal:

[ F_f \cdot d = \frac{1}{2}mv^2 ]

From this relationship, we can solve for stopping distance ( d ):

[ d = \frac{\frac{1}{2}mv^2}{F_f} \

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