Question

On a two-lane road, car A is traveling with a speed of 36 km/h . Two cars B and C approach car A in opposite directions with a speed of 54 km/h each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident? Irrelevant Answers will be reported

Answer 1

Let the car A move rightward.

Speed of car A,

$\longrightarrow\sf{v_A=36\ km\,h^{-1}}$

Let the car B move rightward and car C move leftward, towards the car A, however.

Speed of car B,

$\longrightarrow\sf{v_B=54\ km\,h^{-1}}$

Speed of car C,

$\longrightarrow\sf{v_C=54\ km\,h^{-1}}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(-1.2,0){\sf{B}}\put(48.8,0){\sf{A}}\put(100,0){\sf{C}}\put(0,5){\vector(1,0){22.5}}\put(50,5){\vector(1,0){15}}\put(100,5){\vector(-1,0){22.5}}\put(0,8){ \sf{54\ km\,h^{-1}} }\put(50,8){ \sf{36\ km\,h^{-1}} }\put(80,8){ \sf{54\ km\,h^{-1}} }\multiput(25,-5)(50,0){2}{\put(-7,0){\vector(-1,0){18}}\put(5.5,0){\vector(1,0){19.5}}\put(-4.5,-1){\sf{1\ km}}}\end{picture}$

Speed of each car with respect to car A will be,

$\longrightarrow\sf{v_A-v_A=0\ km\,h^{-1}}$

$\longrightarrow\sf{v_B-v_A=18\ km\,h^{-1}}$

$\longrightarrow\sf{v_C-v_A=90\ km\,h^{-1}}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(-1.2,0){\sf{B}}\put(48.8,0){\sf{A}}\put(100,0){\sf{C}}\put(0,5){\vector(1,0){7.5}}\put(100,5){\vector(-1,0){37.5}}\put(0,8){ \sf{18\ km\,h^{-1}} }\put(80,8){ \sf{90\ km\,h^{-1}} }\multiput(25,-5)(50,0){2}{\put(-7,0){\vector(-1,0){18}}\put(7,0){\vector(1,0){18}}\put(-4.5,-1){\sf{1\ km}}}\end{picture}$

Relative to A, time taken by car C to reach car A is,

$\sf{\longrightarrow t=\dfrac{1}{90}\ h}$

During this time the car B should overtake A by travelling 1 km.

Let minimum acceleration of car B be $\sf{a}$ in $\sf{km\,h^{-1}.}$

By second equation of motion, the distance 1 km to reach A is given as,

$\sf{\longrightarrow 1=18\times\dfrac{1}{90}+\dfrac{1}{2}\times a\left(\dfrac{1}{90}\right)^2}$

$\sf{\longrightarrow 1=\dfrac{1}{5}+\dfrac{1}{2}\times a\times\dfrac{1}{8100}}$

$\sf{\longrightarrow 1=\dfrac{3240+a}{16200}}$

$\sf{\longrightarrow3240+a=16200}$

$\sf{\longrightarrow\underline{\underline{a=12960\ km\,h^{-2}}}}$

$\sf{\longrightarrow a=12960\times\dfrac{1000}{3600\times3600}\ m\,s^{-2}}$

$\sf{\longrightarrow a=12960\times\dfrac{1}{12960}\ m\,s^{-2}}$

$\sf{\longrightarrow\underline{\underline{a=1\ m\,s^{-2}}}}$

Answer 2

$\underline{\underline{\purple{\rm Solution:}}}$

$\rm Speed \: of \: car A(u_A) = 36km/h$

$\rm =36 \times \dfrac{5}{18}m/s \qquad \Bigg( \because 1km/h = \dfrac{5}{18} m/s \Bigg)$

$\rm = 10m/s$

$\rm Speed \: of \: car B \: and \: car C$

$\rm u_B = u_C = 54km/h$

$\rm =54 \times \dfrac{5}{18}m/s$

$\rm = 15m/s$

Relative velocity of Car B w.r.t car A

$\rm u_{BA}= u_B - u_A = 15 - 10 = 5m/s$

Relative velocity of Car C w.r.t car A

$\rm u_{CA}= u_C - u_A = 15 - (-10) = 25m/s$

Distance between car A and car B = 1km = 1000m

Time taken by car C to travel distance AC = 1000m

$\rm Time(t) = \dfrac{Distance}{ Relative \: velocity \: of \: Car C \: w.r.t \: car A }$

$\rm = \dfrac{1000}{25}s = 25s$

Let, car B starts to accelerate with an acceleration a.

$\rm Using \: eq. \: of \: motion,s = ut + \dfrac{1}{2} a {t}^{2}$

$\rm or \qquad \qquad \qquad \quad s = u_{BA}t \dfrac{1}{2} a {t}^{2}$

$\rm \implies 1000 = (5) \times (40) + \dfrac{1}{2} a {(40)}^{2}$

$\rm \implies 1000 = 200 + 800a$

$\rm \implies 1000 - 200 = 800a$

$\rm \implies 800a = 800$

$\rm \implies a = \dfrac{800}{800} = 1{m/s}^2$

$\rm \implies a =1{m/s}^2$

Therefore, car B should accelerate with an acceleration 1m/s²