## Best Answer for this question Your driving the highway late one night at 20m/s when a deer steps onto the road 41m in front.?

This can easily be solved using one of the kinematic equations of motion.

Let's deal with the reaction time first. We know you reaction time is 0.50 s, and the car travels a distance of (20 m/s)(0.5 s) = 10 m in that time, leaving a distance of 41 m - 10 m = 31 m to the deer when you start braking.

Assuming your car decelerates at 10 m/s², the distance it takes to brake can be found using one of the kinematic equations:

u = initial velocity = 20 m/s

v = final velocity = 0 m/s

a = acceleration = -10 m/s²

d = distance = ?

v² = u² + 2ad, and since v = 0, 2ad = -u² so that d = -u²/2a = -(20 m/s²)/(2*-10 m/s²) = 20 m.

So the distance between deer and car at stopping is 31 m - 20 m = 11 m.

And what about the maximum speed? We know u² = -2ad, and you have 31 m to stop in: the maximum initial speed possible is

u² = -2(-10 m/s²)(31 m) = 620 m²/s² and take the square root

u = 24.9 m/s.

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