You and your identical twin are driving identical cars with identical tires down

Question

You and your identical twin are driving identical cars with identical tires down a straight road on a foggy day. You are the more prudent of the two and are driving at a speed of 50 km/hr. Your twin decides to pass you and accelerates to a speed of 70 km/hr, which he maintains as he tries to pass. As his car draws level to yours, each of you sees an overturned truck blocking the road. Both of you apply the brakes at the same instant and begin to skid towards the truck at the same time. You manage to halt a few inches from the truck. Approximately how fast is your twin going when he crashes into the overturned truck?

Explanation / Answer

let the stopping distance for me = dm

retardation from the road = -a                        (retardation is same for both me and my twin)

my initial speed before applying break was = Vmi = 50 km/h

my final speed = Vmf = 0

Using kinematics equation :-

Vmf2 = Vmi2 + 2 a dm

0 = (50)2 + 2 (-a) dm

so dm = 2500/2a                  equation- 1

Consider the motion of my twin

stopping distance for twin = dm     (same as me since truck is same distance away from us Both )

retardation from the road = -a                        (retardation is same for both me and my twin)

initial speed of twin before applying break was = Vti = 70 km/h

twin final speed = Vif = ?

Using kinematics equation :-

Vtf2 = Vti2 + 2 a dm

Vtf2  = (70)2 + 2 (-a) (2500/2a)                      since dm = 2500/2a    

          Vtf2  = 4900 - 2500      = 2400

    Vtf = 48.98 km/h   speed of twin when crashing into truck

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