To avoid an accident a car brakes with a constant deceleration of 6ft/sec^2, producing skid marks?

Question

measuring 75ft before coming to a stop. How fast was the car traveling when the brakes were first applied?

Answer 1

Vf^2 = 2ad + Vi^2

Vf = final velocity
a = acceleration (or deceleration)
d = displacement (or changing position)
Vi = initial velocity (this what we're are going to find)


when the car stops, its final velocity is 0ft/s

what we know:
Vf = 0ft/s
a = -6 ft/s^2 (because the car is slowing down)
d = 75ft

0^2 = 2(-6)(75) + Vi^2
900 = Vi^2
Vi = 30 ft/s

Answer 2

there's a formula that relates the preliminary and extremely final velocities of an merchandise, given the acceleration and distance travelled. we've all of those issues aside from the preliminary velocity, which we are searching for. Very handy! v(very final)^2 = v(preliminary)^2 + 2(acceleration)(distance) v(very final) = 0, acceleration = -forty, distance = one hundred sixty. 0 = v(preliminary)^2 + 2(-forty)(one hundred sixty) v(preliminary)^2 = 12800 v(preliminary) = 113.14 feet/s. This corresponds to a velocity of seventy seven.14 mph. i do no longer understand an outstanding purchase approximately vehicles, yet i think of this automobile had ok brakes :D

Answer 3

a=-6
v=-6t +vo and e= -3t^2+vot
v=0 so t=vo/6 and 75=-3vo^2/36 +vo^2/6
75=vo^2/12 and vo= sqrt(75*12) =30feet/sec