When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (${\text{(}}{{\text{v}}_{\text{0}}}{\text{)}}$ and the braking capacity, or deceleration, ${\text{ - a}}$ that is caused by the braking. Derive and expression for stopping distance of a vehicle in terms of ${{\text{v}}_{\text{0}}}$ and ${\text{a}}$.
Answer
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Hint: Whenever brakes are applied to a moving vehicle, the vehicle doesn’t stop immediately but instead it travels a small distance and then stops. This distance travelled by the moving vehicle before stopping is called stopping distance.
This stopping distance depends on the initial velocity of the vehicle and the braking capacity. Braking capacity is also called as deceleration or negative acceleration or retardation which is in the opposite direction of the movement of the vehicle.
Complete step by step answer:
Initial speed of the moving vehicle ${\text{ = }}{{\text{v}}_{\text{0}}}$
Braking capacity or deceleration ${\text{ = a}}$
Final velocity of the vehicle, v ${\text{ = 0}}$(as the vehicle stops)
Let the distance travelled by the vehicle after applying brakes and when it stops be ‘s’
Then applying the Newton’s third equation of motion we have
${{\text{v}}^{\text{2}}}{\text{ = }}{{\text{v}}_{\text{0}}}^{\text{2}}{\text{ + 2( - a)s}} \\
\Rightarrow {\text{0 = }}{{\text{v}}_{\text{0}}}^{\text{2}}{\text{ - 2as}} \\
\Rightarrow {\text{s = }}\dfrac{{{{\text{v}}_{\text{0}}}^{\text{2}}}}{{{\text{2a}}}}$
This is the expression for stopping distance.
Additional information:
Doubling the initial velocity increases the stopping distance by a factor of 4 for the same deceleration. The stopping distance depends on the vehicle’s speed and weight, in addition to the factors of energy, heat and friction.
Note: Stopping speed is often given as a ${\text{100 - 0kph}}$ distance and is given on a dry pavement. Stopping distance is considered as an important factor in setting speed limits, for example in school zones.
This stopping distance depends on the initial velocity of the vehicle and the braking capacity. Braking capacity is also called as deceleration or negative acceleration or retardation which is in the opposite direction of the movement of the vehicle.
Complete step by step answer:
Initial speed of the moving vehicle ${\text{ = }}{{\text{v}}_{\text{0}}}$
Braking capacity or deceleration ${\text{ = a}}$
Final velocity of the vehicle, v ${\text{ = 0}}$(as the vehicle stops)
Let the distance travelled by the vehicle after applying brakes and when it stops be ‘s’
Then applying the Newton’s third equation of motion we have
${{\text{v}}^{\text{2}}}{\text{ = }}{{\text{v}}_{\text{0}}}^{\text{2}}{\text{ + 2( - a)s}} \\
\Rightarrow {\text{0 = }}{{\text{v}}_{\text{0}}}^{\text{2}}{\text{ - 2as}} \\
\Rightarrow {\text{s = }}\dfrac{{{{\text{v}}_{\text{0}}}^{\text{2}}}}{{{\text{2a}}}}$
This is the expression for stopping distance.
Additional information:
Doubling the initial velocity increases the stopping distance by a factor of 4 for the same deceleration. The stopping distance depends on the vehicle’s speed and weight, in addition to the factors of energy, heat and friction.
Note: Stopping speed is often given as a ${\text{100 - 0kph}}$ distance and is given on a dry pavement. Stopping distance is considered as an important factor in setting speed limits, for example in school zones.
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