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An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be
- a)60 m
- b)40 m
- c)20 m
- d)80 m
Correct answer is option 'D'. Can you explain this answer?
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An automobile travelling with a speed of 60 km/h, can brake to stop wi...
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An automobile travelling with a speed of 60 km/h, can brake to stop wi...
- Analysis:
An important concept to consider in this scenario is the relationship between speed and stopping distance. When a car is traveling at a higher speed, it will require a longer distance to come to a complete stop due to the increased momentum.
- Given Data:
Initial speed of the car, V1 = 60 km/h
Stopping distance at this speed, d1 = 20m
- Calculation:
Using the formula for stopping distance:
d = (V^2) / 2μg
where,
d is the stopping distance,
V is the initial velocity,
μ is the coefficient of friction,
and g is the acceleration due to gravity.
- Calculation for V1 = 60 km/h:
Converting the speed to m/s:
V1 = 60 km/h * (1000 m / 3600 s) = 16.67 m/s
Plugging in the values:
20 = (16.67^2) / 2μg
μg = (16.67^2) / 40
μg = 6.945
- Calculation for V2 = 120 km/h:
Converting the speed to m/s:
V2 = 120 km/h * (1000 m / 3600 s) = 33.33 m/s
Using the same coefficient of friction:
d2 = (33.33^2) / 2*6.945
d2 = 80 m
- Conclusion:
The stopping distance for a car traveling at 120 km/h is 80 meters, which is twice the stopping distance when the car is traveling at 60 km/h (20 meters). Therefore, the correct answer is option D) 80 meters.
An important concept to consider in this scenario is the relationship between speed and stopping distance. When a car is traveling at a higher speed, it will require a longer distance to come to a complete stop due to the increased momentum.
- Given Data:
Initial speed of the car, V1 = 60 km/h
Stopping distance at this speed, d1 = 20m
- Calculation:
Using the formula for stopping distance:
d = (V^2) / 2μg
where,
d is the stopping distance,
V is the initial velocity,
μ is the coefficient of friction,
and g is the acceleration due to gravity.
- Calculation for V1 = 60 km/h:
Converting the speed to m/s:
V1 = 60 km/h * (1000 m / 3600 s) = 16.67 m/s
Plugging in the values:
20 = (16.67^2) / 2μg
μg = (16.67^2) / 40
μg = 6.945
- Calculation for V2 = 120 km/h:
Converting the speed to m/s:
V2 = 120 km/h * (1000 m / 3600 s) = 33.33 m/s
Using the same coefficient of friction:
d2 = (33.33^2) / 2*6.945
d2 = 80 m
- Conclusion:
The stopping distance for a car traveling at 120 km/h is 80 meters, which is twice the stopping distance when the car is traveling at 60 km/h (20 meters). Therefore, the correct answer is option D) 80 meters.
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An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will bea)60 mb)40 mc)20 md)80 mCorrect answer is option 'D'. Can you explain this answer?
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