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A driver is traveling at 30.0 m/s when he sees a moose crossing the road 80.0 m ahead. The moose becomes distracted and stops in the middle of the road. If the driver of the car slams on the brakes, what is the minimum constant acceleration he must undergo to stop short of the moose and avert an accident?

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okpalawalter8

Answer:

The value is [tex]a = -5.625 \ m/s^2[/tex]

Explanation:

From the question we are told that

The speed of the driver is [tex]u = 30.0 \ m/s[/tex]

The distance of the car from the moose is [tex]s= 80 \ m[/tex]

Generally from the third equation of motion we have that

[tex]v ^2 = u ^2 + 2as[/tex]

Here  v =  0 \  m/s

So

     [tex]0  =  30 ^2  +  2 * a * 80[/tex]

       [tex]a =  -5.625 \ m/s^2[/tex]

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