Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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.

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.