Newtons Laws EX 17

$v_{0}=90{\frac {km}{h}}=90{\frac {km{\frac {1000m}{1km}}}{h{\frac {3600s}{1h}}}}=90\times {\frac {1000m}{3600s}}=90\times {\frac {1m}{3.6s}}=25{\frac {m}{s}}$

$v=0$

$\Delta x=75cm=0.75m$

$v^{2}-{v_{o}}^{2}=2a\Delta x$

$0-(25m/s)^{2}=2a(0.75m)$

$a=-416.7m/s^{2}=-420m/s^{2}$ (Note that we need to round off to two significant digits.)

Therefore the magnitude of the net force acting on the driver is

$F_{net}=420\times 75=31500N=32000N$ (Note that we need to round off to two significant digits.)

Further note that the magnitude of the force is always positive. The force acts in the direction opposite to the velocity as indicated by the negative sign of the acceleration. Note the enormity of this force which is largely a consequence of high speed of the vehicle before the crash.