Menu Close

On-a-two-lane-road-car-A-is-travelling-with-a-speed-of-36km-h-Two-cars-B-and-C-approach-A-in-opposite-directions-with-a-speed-of-54km-h-each-At-a-certain-instant-when-the-distance-AB-is-equal-to-AC-




Question Number 29359 by NECx last updated on 08/Feb/18
On a two lane road,car A is  travelling with a speed of 36km/h.  Two cars B and C approach A in  opposite directions with a speed of  54km/h each.At a certain instant,  when the distance AB is equal  to AC,both being 1km,B decides  to overtake A before C does.What  minimum acceleration of carB  is required to avoid an accident?
$${On}\:{a}\:{two}\:{lane}\:{road},{car}\:{A}\:{is} \\ $$$${travelling}\:{with}\:{a}\:{speed}\:{of}\:\mathrm{36}{km}/{h}. \\ $$$${Two}\:{cars}\:{B}\:{and}\:{C}\:{approach}\:{A}\:{in} \\ $$$${opposite}\:{directions}\:{with}\:{a}\:{speed}\:{of} \\ $$$$\mathrm{54}{km}/{h}\:{each}.{At}\:{a}\:{certain}\:{instant}, \\ $$$${when}\:{the}\:{distance}\:{AB}\:{is}\:{equal} \\ $$$${to}\:{AC},{both}\:{being}\:\mathrm{1}{km},{B}\:{decides} \\ $$$${to}\:{overtake}\:{A}\:{before}\:{C}\:{does}.{What} \\ $$$${minimum}\:{acceleration}\:{of}\:{carB} \\ $$$${is}\:{required}\:{to}\:{avoid}\:{an}\:{accident}? \\ $$
Answered by 33 last updated on 08/Feb/18
  v_(rel C )  = 15 −(−10 )= 25 m/s  v_(rel B ) = 15 − 10  = 5 m/s  B must reach A just before C  reaches to avoid accident.   let t_C  ⋍ t_B   t_C  = ((1000)/(25)) = 40 s            (S_(rel) /v_(rel) )  using second equation of motion  1000 = 5(40) + (1/2)(a)(40^2 )  ⇒ a = 1 m/s^2  if they reach A  simultaneously.  ⇒ a ⪈ 1 m/s^2
$$ \\ $$$${v}_{{rel}\:{C}\:} \:=\:\mathrm{15}\:−\left(−\mathrm{10}\:\right)=\:\mathrm{25}\:{m}/{s} \\ $$$${v}_{{rel}\:{B}\:} =\:\mathrm{15}\:−\:\mathrm{10}\:\:=\:\mathrm{5}\:{m}/{s} \\ $$$${B}\:{must}\:{reach}\:{A}\:{just}\:{before}\:{C} \\ $$$${reaches}\:{to}\:{avoid}\:{accident}. \\ $$$$\:{let}\:{t}_{{C}} \:\backsimeq\:{t}_{{B}} \\ $$$${t}_{{C}} \:=\:\frac{\mathrm{1000}}{\mathrm{25}}\:=\:\mathrm{40}\:{s}\:\:\:\:\:\:\:\:\:\:\:\:\left({S}_{{rel}} /{v}_{{rel}} \right) \\ $$$${using}\:{second}\:{equation}\:{of}\:{motion} \\ $$$$\mathrm{1000}\:=\:\mathrm{5}\left(\mathrm{40}\right)\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\mathrm{40}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{a}\:=\:\mathrm{1}\:{m}/{s}^{\mathrm{2}} \:{if}\:{they}\:{reach}\:{A} \\ $$$${simultaneously}. \\ $$$$\Rightarrow\:{a}\:\gneq\:\mathrm{1}\:{m}/{s}^{\mathrm{2}} \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *