Answered

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A drag racer starts from rest and accelerates at 43 m/s2. After 2.9 seconds of this
constant acceleration, how far down the track is the dragster? (Answer in meters)

Sagot :

confucius

Answer:

180 meters

Explanation:

We know initial position and velocity of the dragster are both zero. We only know acceleration (43 m/s^2) and need to find out how far the dragster moves after 2.9 seconds.

x initial = 0

velocity initial = 0

t=2.9s

a= 43m/s^2

Plug into

[tex]x_{final} =x_{initial} +v_{initial}t+\frac{1}{2} at^2[/tex]

[tex]x_{final} =0} +0t+\frac{1}{2} (43 m/s^2)(2.9s)^2[/tex]

[tex]x_{final} =\frac{1}{2} (43 m/s^2)(8.41s^2)[/tex]

[tex]x_{final}=180.815m = 180m[/tex]

180 meters or 1.8 x 10^2 meters is the answer accounting for 2 sig figs.

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