You're driving at speed v_0 when you spot a stationary moose on the road, a distance d ahead. Find an expression for the magnitude of the acceleration you need if you're to stop before hitting the moose. Please explain thanks.

The formula that is applicable in this problem is the formula in kinematics vf^2 = vi^2 + 2ad
Initial velocity = Vi Final velocity = 0 (as the car stops) Distance travelled = d
vf^2 = vi^2 + 2ad (0)^2 = (Vi)^2 + 2ad 0 = (Vi)^2 + 2ad 2ad = -(Vi)^2 a = - (Vi)^2 / (2d)

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aboujokhyamal aboujokhyamal · Answer 2 · 2019-08-09T15:08:04+02:00

In case your car brakes at a constant rate (meaning your deceleration is constant), then you would have to decelerate at [tex]\frac{V^2}{2 \cdot d}[/tex].

Further explanation

When bodies move they have what is called "linear momentum". Because of Newton's laws of motion, when a force goes against the body's movement, it will tend to slow it down (thus making its momentum smaller). When we apply these concepts to a car, it's easier to understand, since if a car is going at a certain speed and it breaks, due to the force applied on the brakes (which goes against the car's movement) the car slows down.

A particular issue, is how the car breaks (sometimes the car can break harder than at other moments) and this will have an important impact in our solution, however due to the simplicity of the problem we will assume that the deceleration of the car is constant (in case this is not assumed, then we're missing important data to solve the problem). Based on these assumption, we can easily compute the stopping distance from the following equation:

[tex]V_f ^2 = V_0 ^2 + 2 \cdot a \cdot d[/tex]

Where [tex]V_f[/tex] is the final velocity, [tex]V_0[/tex] is the initial velocity of the car, a is the car's acceleration, and d is the stopping distance. Since we want to stop before hitting the moose, the final velocity must be 0, and so we can solve for the deceleration as:

[tex]0= V_0 ^2 + 2 \cdot a \cdot d[/tex]

[tex]a = - \frac{V_0 ^2}{2\cdot d} [/tex]

As an example, we can take common values to use our equation. Let's say we're driving at 30 mph (which translates to around 13.5 meters per second), and we see a moose 100 meters ahead. Then our deceleration would be:

[tex]a = - \frac{(13.5) ^2}{2\cdot 100} = -0.91 \frac{m}{s^2}[/tex]

Which is a not that high deceleration.

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Keywords

Kinematics, velocity, speed, stopping distance, deceleration.