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To practice Problem-Solving Strategy 2.1 for constant acceleration problems. A car is traveling at a constant velocity of magnitude v0 when the driver notices a garbage can on the road in front of him. At that moment, the distance between the garbage can and the front of the car is d. A time t after noticing the garbage can, the driver applies the brakes and slows down at a constant rate before coming to a halt just before the garbage can. What is the magnitude of the car's acceleration after the brakes are applied?

Respuesta :

Answer:[tex]a=\frac{v_0^2}{2(d-v_0t)}[/tex]

Explanation:

Given

Velocity of car is [tex]v_0[/tex]

distance between car and can is d

reaction time is t

Distance traveled in reaction time

let say [tex]d_0[/tex]

[tex]d_0=v_0t[/tex]

Remaining distance[tex]=d-d_0[/tex]

Now final velocity of  car will be zero after applying brakes

Using equation of motion

[tex]v^2-u^2=2as[/tex]

[tex]0-v_0^2=2\times (-a)\times (d-d_0)[/tex]

[tex]a=\frac{v_0^2}{2(d-v_0t)}[/tex]

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