06-28-2023, 07:38 AM
Only because I like pedantic arguments like this, I'll "weigh" in.
The simplest way to think about it is in terms of deceleration. Since both the motorcycle and the car start out at the same velocity, if the motorcycle travels less distance, it's because it decelerated to zero quicker. Why would that be? Since F=m*a, a=F/M where F is frictional and aerodynamic force in the rearward direction. If that force were identical between the car and the motorcycle, clearly the motorcycle would decelerate quicker since its mass is less.
The frictional and aerodynamic force for the car is certainly going to be more than for a motorcycle of course, but not enough in comparison with the mass difference. So the car takes longer to decelerate to zero. Same holds true for m-in-sc's train example.
There are certainly examples though of vehicles that weigh more, but travel shorter distances from speed than some other vehicle that weighs less. For most wheeled vehicles though, the heavier ones will decelerate more slowly due to frictional and drag forces.
The simplest way to think about it is in terms of deceleration. Since both the motorcycle and the car start out at the same velocity, if the motorcycle travels less distance, it's because it decelerated to zero quicker. Why would that be? Since F=m*a, a=F/M where F is frictional and aerodynamic force in the rearward direction. If that force were identical between the car and the motorcycle, clearly the motorcycle would decelerate quicker since its mass is less.
The frictional and aerodynamic force for the car is certainly going to be more than for a motorcycle of course, but not enough in comparison with the mass difference. So the car takes longer to decelerate to zero. Same holds true for m-in-sc's train example.
There are certainly examples though of vehicles that weigh more, but travel shorter distances from speed than some other vehicle that weighs less. For most wheeled vehicles though, the heavier ones will decelerate more slowly due to frictional and drag forces.