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During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a 3 second time period. What was his average acceleration over that 3 second period?

Respuesta :

lublana

Answer:

[tex]6.67ft/s^2[/tex]

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=[tex]\frac{v-u}{t}{t}[/tex]

Using the formula

Average acceleration,a=[tex]\frac{38-18}{3}ft/s^2[/tex]

Average acceleration,a=[tex]\frac{20}{3}ft/s^2[/tex]

Average acceleration,a=[tex]6.67ft/s^2[/tex]

Hence, the average acceleration=[tex]6.67ft/s^2[/tex]

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