Time taken to stop: 3.0 s
Explanation:
The motion of the car is a uniformly accelerated motion, therefore we can use the following suvat equation:
[tex]s=vt-\frac{1}{2}at^2[/tex]
where
s is the displacement
v is the final velocity
t is the time
a is the acceleration
For the car in this problem, we have:
s = 9.0 m (displacement)
v = 0 (the car comes to a stop)
[tex]a=-2.0 m/s^2[/tex] (deceleration)
t is time
And solving for t, we find the time the car rakes to stop:
[tex]t=\sqrt{-\frac{2s}{a}}=\sqrt{\frac{2(9.0)}{-2.0}}=3.0 s[/tex]
Learn more about accelerated motion:
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