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a bicyclist approaches the credt of a hill at 4.5 m/s. she accelerates down the hill at a rate of 0.40 m/s^2 for 12 s. how far does she move down the hill during this time interval?​

Sagot :

quasarJose

Answer:

25.3m

Explanation:

Given parameters:

Final velocity   =  4.5m/s

Acceleration  = 0.4m/s²  

Time taken  = 12s

Unknown:

Distance she moved down hill  = ?

Solution:

To solve this problem, we use the expression below:

              v² = u² + 2aS

v is the final velocity

u is the initial velocity

a is the acceleration

S is the distance

             4.5² = 0²  + (2 x 0.4 x S)

              20.25  = 0.8S

                S = 25.3m

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