Acar traveling at 22.4 m/s skids to a stop in 2.25s. determine the skidding distance of the car

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A car traveling at 22.4 m/s skids to a stop in 2.25 s. Determine the skidding distance of the car

We have the following data:

a (acceleration) = ? (in m/s²)

t (time) = 2.25 s

Vf (final velocity) = 22.4 m/s

Vi (initial velocity) = 0 m/s

We apply the data to the formula of the hourly function of the velocity, let's see:

*** The distance traveled ?  

We have the following data:  

Vi (initial velocity) = 0 m/s  

t (time) = 2.25 s  

a (average acceleration) = 9.955 m/s²  

d (distance interval) = ? (in m)

By the formula of the space of the Uniformly Varied Movement, it is:

*** Another way to solve:

We have the following data:  

Vf (final velocity) = 22.4 m/s

Vi (initial velocity) = 0 m/s  

a (average acceleration) = 9.955 m/s²  

d (distance interval) = ? (in m)

Since we do not need to know the time elapsed during the movement, we apply the data of the question to the Equation of Torricelli, let us see:

________________________

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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