Question

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?

Answer 1

Total distance travelled by the car is 'd' 
distance trveled before the brakes were applied = v_o * t 
distance travld with brakes = d - v_o*t 
applying the formula: v^2 - u^2 = -2 a * s 
=> 0 - v_o^2 = -2 * a_x * (d- v_o*t) 
=> a_x = (v_o^2)/ ( 2 (d-v_o*t)

Answer 2

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

$v_f^2 - v_i^2 = 2 a d$

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

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