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velikii [3]
2 years ago
14

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?
Physics
2 answers:
slava [35]2 years ago
7 0
Total distance travelled by the car is 'd' 
<span>distance trveled before the brakes were applied = v_o * t </span>
<span>distance travld with brakes = d - v_o*t </span>
<span>applying the formula: v^2 - u^2 = -2 a * s </span>
<span>=> 0 - v_o^2 = -2 * a_x * (d- v_o*t) </span>
<span>=> a_x = (v_o^2)/ ( 2 (d-v_o*t)</span>
yanalaym [24]2 years ago
4 0

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)}

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