here,
the initial speed of car , u = 44.7 m/s
the final speed of car , v = 0 m/s^2
acceleration of car , a = - 20 m/s^2
let the stopping distance be s
using seccond equation of motion
v^2 - u^2 = 2 * a * s
0^2 - 44.7^2 = 2 * ( -20) * s
solving for s
s = 50 m
the stopping distance for car is 50 m
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During a very quick stop, a car
decelerates at 7.8 m/s2. Assume the forward motion of
the car corresponds to a positive direction for the rotation of the
tires (and that they do not slip on the pavement).
Randomized Variablesat = 7.8
m/s2
r = 0.29 m
ω0 = 93 rad/s
Part (a) What is the angular acceleration of
its tires in rad/s2, assuming they have a radius of 0.29
m and do not slip on the pavement?
Part (b)...
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