The car collision accident analysis problem, as Figure below: The driver of a 10

Question

The car collision accident analysis problem, as Figure below: The driver of a 1000 kg car traveling on the interstate at 65 MPH (you need to convert MPH into m/s, 1 mile = 1609 m, 1 hour =3600 s) applies his bake to avoid hitting a second car in front of him which has fully stopped. After the brakes are applied, a constant friction force of 7800 n acts on his car. At what minimum distance should the brakes be applied to avoid a collision with the other car? If the distance between the cars is initially on 30.0 m, at what speed would the collision occur? A car of mass 1020.0 kg accelerates from rest with a constant power output of 130.0 hp (1 hp = 746 w). Neglecting air resistance, what is the speed of the car after 6.55 s.

Explanation / Answer

(1) (a) Convert MPH into m/s -

So, 65 MPH = (65x1609) / 3600 = 29 m/s.

Now apply work energy theorem -

W = (1/2)m(Vf^2 - Vi^2)

=> Ff * delta x = (1/2)m(0 - Vi^2)

=> 7800*delta x = 0.5*1000*29^2

=> delta x = 53.9 m

(b) From above -

- Ff * delta x = (1/2)m(Vf^2 - Vi^2)

Now, multiply by 2/m on both sides -

Vf^2 = Vi^2 - (2/m)*Ff*delta x

=> Vf^2 = 29^2 - (2/1000)*7800*30 = 373

=> Vf = 19.3 m/s

(2) Here, P = F v

P = m v dv/dt

=> m v dv = P dt

=> m v^2 /2 = P t

{ P = 130 hp = 130 x 746 Watt }

Now, put the values-

1020 v^2 /2 = (130 x 746) (6.55 - 0)

=> v = 35.3 m/s

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