starting from rest the road runner acceerates at 2.8m/s2 for 9.0seconfs how far does the road runner travels during the nine second time interval

Respuesta :

Answer:

The roadrunner traveled 113.4 m in 9 seconds.

Explanation:

Given:

Initial velocity of road runner (u) = 0 m/s 9 (Rest)

Acceleration of road runner (a) = 2.8 m/s²

Time taken by road runner (t) = 9.0 s

Now, displacement of the road runner can be determined using Newton's equation of motion.

In order to find the displacement, we need to use the equation of motion that relates displacement, initial velocity, acceleration and time.

Therefore, the equation of motion is given as:

[tex]S=ut+\frac{1}{2}at^2\\\\S\to displacement[/tex]

Plug in the given values and solve for 'S'. This gives,

[tex]S=0+\frac{1}{2}(2.8)(9^2)\\\\S=1.4\times 81\\\\S=113.4\ m[/tex]

Therefore, the roadrunner traveled 113.4 m in 9 seconds.

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