Enterprising students set an enormous slip-n-slide (a plastic sheetcovered in water to reduce friction) on flat ground. If the slip-n-slideis 250 m long, how small does the average acceleration have to be fora student starting at 5 m/s to slide to the end

a = -0.05 m/s² (negative sign shows deceleration)

Explanation:

In order, to find out the minimum average acceleration for a student starting at 5 m/s to slide to the end, we can use 3rd equation of motion. 3rd equation of motion is given as follows:

2as = Vf² - Vi²

where,

a = minimum acceleration required = ?

s = minimum distance covered = 250 m

Vf = Final Speed = 0 m/s (for minimum acceleration the student will barely cover 250 m and then stop)

Vi = Initial Velocity = 5 m/s

Therefore,

2a(250 m) = (0 m/s)² - (5 m/s)²

a = - (25 m²/s²)/(500 m)

a = -0.05 m/s² (negative sign shows deceleration)

i belive it is d. both a and c

hope this

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