Answer to Question #224634 in Mechanical Engineering for Papi Chulo

Question

A woman sitting in a stationary car notices a man cycling past her at 40 km/hr. Five

seconds after he passes her car, she starts following after him. What is the shortest

possible distance in which she will catch up to the cyclist if he maintains speed and

she obeys the speed limit of 70 km/hr? The car has a maximum acceleration of 3.5m.s2

Answer 1

v=u+at

u=0, v=70km/hr÷3.6=19.44 m/s

t=v÷a=19.44÷3.5=5.55s

d=s*t

s=40km/hr÷3.6=11.11m/s

d=11.11m/s*5=55.55m

T=$\frac{distance}{relative speed}=\frac{55.55m}{19.44-11.11}=6.67s$

s₁=$\frac{d}{5.55s}$

s₂= 11.11m/s

T=$\frac{distance}{relative density}$

$6.67=\frac{55.55}{(\frac{d}{5.55}-11.11)}$

$6.67(\frac{d}{5.55}-11.11)=55.55$

$\frac{d}{5.55}=\frac{55.55}{6.67}+11.11$

d=19.44*5.55

=107.89m