Question: Two cars are travelling on a straight road. The first car is travelling at 55 km per hour. The second car is 1.5 m behind it and travelling at 52 km per hour. The first car is 2 m long and 1.5 m wide. What distance does the second car travel to overtake the first car, assuming that the second car moves 1 metre aside the first car and accelerates to 60 km per hour at a rate of 0.5 km per second?

Posted by: Tatum S. on 05.11.2019

It is presumed that there is sufficient side clearance between the two cars so that the second car can safely overtake the first car.

The distance between the two cars at start = Lead of  first car over the second + length of the first car

=  1.5  +2 = 3.5  metres( As the length of the second car was not given  the same is not considered here.)

Speed of first car                                          =   55Km/hr

Initial speed of second  car                        =  52km/hr

The acceleration of second car                  = 0.5km/hr per sec

The second car accelerates to a speed of 60km per hr from an initial speed of 52km/hr in 16 seconds and thereafter maintains a constant speed of 60km/hr. through out.

The speed of the 2nd car at any instant of time’  t’ seconds during acceleration   = (52+ 0.5t) km/hr

Therefore the average speed of the car during any instant ‘t’=  (52+52+0.5t)/2=  (52+0.25t) km/hr

Relative speed of 2 cars at any time’ t’ during acceleration= (52+0.25t-55)= (0.25t-3 )km/hr

Let the time taken for the second car to overtake first car =  t seconds

Time ‘t’ taken for the second car to overtake the first car= Initial distance during start/Relative speed

That is   t   ={ 3.5*10^(-)3}/(0.25t-3)*(1/3600) ( converting speeds  into km/s and distance into kms.)

that is ,t=       3.5*3.6/(0.25t-3) . By  cross multiplying we get ( 0.25t-3)*t= 3.5*3.6 =12.6

that is 0.25t^2-3t=12.6

that is 0.25 t^2  -3t-12.6=0 ; or t^2-12t=50.4=0  this is a quadratic equation  in ‘t’ and solving for ‘t’ we get

t=  {12+√345.6}/2 and  {12-√345.6}/2. Rejecting the negative value we get

t=15.295 seconds.

There fore the distance travelled  by the second car in overtaking  the first car=

Average speed  * time taken= {52+0.25*15.295}*15.295/3600=    0.23717 km=237.17 metres.

The second car cannot overtake the first car by moving aside only by 1 m as the first car is 1.5 m wide. In oerder to overtake it, the second car has to move aside at least by 2 m.

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