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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

Da 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?

Verified Answer :

Answer:

\(a = \frac{v_o^2}{2(d - v_ot)}\)

Explanation:

distance moved by the car while the driver reacted to apply brakes is given as

\(d_1 = v_o t\)

now the distance of the car from the position of garbage is given as

\(d_2 = d - v_o t\)

now if driver applies rakes due to which car decelerates uniformly and comes to rest

so we have

\(v_f^2 - v_i^2 = 2 a d\)

\(0 - v_o^2 = 2(-a)(d - v_ot)\)

\(a = \frac{v_o^2}{2(d - v_ot)}\)

onyebuchinnaji

The magnitude of the acceleration of the car is \(a = \frac{v_0^2}{2d- 2v_0t}\)

The given parameters;

  • the constant velocity of the car, = \(v_0\)
  • the distance between the car and the garbage = d
  • the time when the driver noticed the garbage = t

To find:

  • the magnitude of the car's acceleration after the brakes have been applied

The distance traveled by the  car  before the driver discovers the garbage;

\(d_1 = v_0t\)

The new distance between the driver and garbage;

\(d_2 = d - v_0t\)

The magnitude of the acceleration of the car is calculated as;

\(v_f^2 = v_0^2 + 2ad_2\)

when the driver reaches the final distance (\(d_2\)) after the breaks are  applied, the final velocity (\(v_f\)) of the car will be zero.

\(0 = v_0^2 + 2ad_2\\\\2ad_2 = v_0^2\\\\a= \frac{v_0^2}{2d_2}\\\\recall, \ d_2 = d- v_0t\\\\a = \frac{v_0^2}{2d - 2v_0t}\)

Thus, the magnitude of the acceleration of the car is \(a = \frac{v_0^2}{2d- 2v_0t}\)

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