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Question Number 30613 by NECx last updated on 23/Feb/18
A car negotiates a bend of radius  20m with an acceleration of   12m/s^2 .What is the maximum  speed the car can attain without  skidding?
$${A}\:{car}\:{negotiates}\:{a}\:{bend}\:{of}\:{radius} \\ $$$$\mathrm{20}{m}\:{with}\:{an}\:{acceleration}\:{of}\: \\ $$$$\mathrm{12}{m}/{s}^{\mathrm{2}} .{What}\:{is}\:{the}\:{maximum} \\ $$$${speed}\:{the}\:{car}\:{can}\:{attain}\:{without} \\ $$$${skidding}? \\ $$
Commented by NECx last updated on 24/Feb/18
please help
$${please}\:{help} \\ $$
Answered by mrW2 last updated on 25/Feb/18
v_(max) =(√(ar))=(√(12×20))=15.5m/s=56 km/h
$${v}_{{max}} =\sqrt{{ar}}=\sqrt{\mathrm{12}×\mathrm{20}}=\mathrm{15}.\mathrm{5}{m}/{s}=\mathrm{56}\:{km}/{h} \\ $$
Commented by NECx last updated on 25/Feb/18
can I know how you got the  formula please?
$${can}\:{I}\:{know}\:{how}\:{you}\:{got}\:{the} \\ $$$${formula}\:{please}? \\ $$
Commented by mrW2 last updated on 25/Feb/18
the car can reach an acceleration from  a=12 m/s^2 , that means a friction  coeefficient between tires and ground  is μ=(a/g)=1.2. when the car goes  into the curve and its speed increases,  a part of the friction should be left for  the centrifugal force, that means  with increasing speed the acceleration  should be decreased. when the car  obtains the max. speed the acceleration  should be zero and the complete  frictionforce can be avaiable for the  centrifugal force, i.e.  m(v_(max) ^2 /r)=μmg=ma_(max)   ⇒v_(max) =(√(ra_(max) ))=(√(20×12))=15.5 m/s
$${the}\:{car}\:{can}\:{reach}\:{an}\:{acceleration}\:{from} \\ $$$${a}=\mathrm{12}\:{m}/{s}^{\mathrm{2}} ,\:{that}\:{means}\:{a}\:{friction} \\ $$$${coeefficient}\:{between}\:{tires}\:{and}\:{ground} \\ $$$${is}\:\mu=\frac{{a}}{{g}}=\mathrm{1}.\mathrm{2}.\:{when}\:{the}\:{car}\:{goes} \\ $$$${into}\:{the}\:{curve}\:{and}\:{its}\:{speed}\:{increases}, \\ $$$${a}\:{part}\:{of}\:{the}\:{friction}\:{should}\:{be}\:{left}\:{for} \\ $$$${the}\:{centrifugal}\:{force},\:{that}\:{means} \\ $$$${with}\:{increasing}\:{speed}\:{the}\:{acceleration} \\ $$$${should}\:{be}\:{decreased}.\:{when}\:{the}\:{car} \\ $$$${obtains}\:{the}\:{max}.\:{speed}\:{the}\:{acceleration} \\ $$$${should}\:{be}\:{zero}\:{and}\:{the}\:{complete} \\ $$$${frictionforce}\:{can}\:{be}\:{avaiable}\:{for}\:{the} \\ $$$${centrifugal}\:{force},\:{i}.{e}. \\ $$$${m}\frac{{v}_{{max}} ^{\mathrm{2}} }{{r}}=\mu{mg}={ma}_{{max}} \\ $$$$\Rightarrow{v}_{{max}} =\sqrt{{ra}_{{max}} }=\sqrt{\mathrm{20}×\mathrm{12}}=\mathrm{15}.\mathrm{5}\:{m}/{s} \\ $$
Commented by mrW2 last updated on 25/Feb/18
Commented by mrW2 last updated on 25/Feb/18
at point 1:  f=ma_(max)   at point 2:  f=(√((ma)^2 +(m(v^2 /r))^2 ))=m(√(a^2 +((v^2 /r))^2 ))  at point 3:  f=m(v_(max) ^2 /r)
$${at}\:{point}\:\mathrm{1}: \\ $$$${f}={ma}_{{max}} \\ $$$${at}\:{point}\:\mathrm{2}: \\ $$$${f}=\sqrt{\left({ma}\right)^{\mathrm{2}} +\left({m}\frac{{v}^{\mathrm{2}} }{{r}}\right)^{\mathrm{2}} }={m}\sqrt{{a}^{\mathrm{2}} +\left(\frac{{v}^{\mathrm{2}} }{{r}}\right)^{\mathrm{2}} } \\ $$$${at}\:{point}\:\mathrm{3}: \\ $$$${f}={m}\frac{{v}_{{max}} ^{\mathrm{2}} }{{r}} \\ $$
Commented by NECx last updated on 25/Feb/18
wow... Thanks so much.  I really love it when you explain  with the aid of diagrams.  Thanks once again.
$${wow}…\:{Thanks}\:{so}\:{much}. \\ $$$${I}\:{really}\:{love}\:{it}\:{when}\:{you}\:{explain} \\ $$$${with}\:{the}\:{aid}\:{of}\:{diagrams}. \\ $$$${Thanks}\:{once}\:{again}. \\ $$
Commented by mrW2 last updated on 27/Feb/18
Does my answer match with that in  your book?
$${Does}\:{my}\:{answer}\:{match}\:{with}\:{that}\:{in} \\ $$$${your}\:{book}? \\ $$
Commented by NECx last updated on 06/Mar/18
yes sir.
$${yes}\:{sir}. \\ $$

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