- #1
- 858
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A ballistics-esque question. Suppose we are driving at ##x \text{ m/s}## and we have to suddenly brake. For students in driving schools they are told something along the lines of "during the braking, the body experiences ##y \text{ kg}## of force" or "the mass of the body is much higher than at rest". Eventually they point out the dangers of driving without a seatbelt and that's all fine and dandy, but the technicalities are a bit odd to me. How do they come up with some specific numbers?
Suppose we drive at ##30 \text{ m/s} ##. If the driver's mass is ##80\text{ kg}##, then his kinetic energy would be computable by ##E = \frac{mv^2}{2}## yielding ##36 \text{ kJ} ##. How does one convert this to units of mass of an invisible body that is exerted on the driver during braking?
Alternatively we could also make use of
[tex]
d = \frac{v^2 - v_0 ^2}{2a}
[/tex]
where ##v## is the terminal velocity, ##v_0## is initial velocity, ##d## is distance and ##a## is acceleration. Suppose we have to come to a full stop in some ##20\text{ m}##, thus requiring an acceleration of ##-22.5 \text{ m/s}^2 ##. So the body would experience a bit more than ##2\text{ g}## of force (I'm not sure if this expression makes sense).
From the above it's not clear to me how one comes up with expressions of the form "at ##x \text{ km/h}## the driver weighs ##y## times more than at rest". Are there other lines of computations to be considered? What do they mean when they say "the body weighs ##y## units at velocity ##z## units" ?
Suppose we drive at ##30 \text{ m/s} ##. If the driver's mass is ##80\text{ kg}##, then his kinetic energy would be computable by ##E = \frac{mv^2}{2}## yielding ##36 \text{ kJ} ##. How does one convert this to units of mass of an invisible body that is exerted on the driver during braking?
Alternatively we could also make use of
[tex]
d = \frac{v^2 - v_0 ^2}{2a}
[/tex]
where ##v## is the terminal velocity, ##v_0## is initial velocity, ##d## is distance and ##a## is acceleration. Suppose we have to come to a full stop in some ##20\text{ m}##, thus requiring an acceleration of ##-22.5 \text{ m/s}^2 ##. So the body would experience a bit more than ##2\text{ g}## of force (I'm not sure if this expression makes sense).
From the above it's not clear to me how one comes up with expressions of the form "at ##x \text{ km/h}## the driver weighs ##y## times more than at rest". Are there other lines of computations to be considered? What do they mean when they say "the body weighs ##y## units at velocity ##z## units" ?