The new stopping distance is 152.6 m
Explanation:
The motion of the car is a uniformly accelerated motion, so we can use the following SUVAT equation:
[tex]v^2-u^2=2as[/tex]
where in the initial situation, we have:
v = 0 is the final velocity of the car
u = 43 km/h = 11.9 m/s is the initial velocity
s = 17 m is the stopping distance
a is the acceleration
Solving for a, we find the acceleration:
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-(11.9)^2}{2(17)}=-4.2 m/s^2[/tex]
The acceleration depends only on the sliding friction: this means that it is constant also in the second situation, where we have
u = 129 km/h = 35.8 m/s
So we can use the same equation again to find s, the new stopping distance:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(35.8)^2}{2(-4.2)}=152.6 m[/tex]
Learn more about accelerated motion:
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