A car accelerates uniformly over a distance of 240 meters in 8 seconds, resulting in an average velocity of 30 m/s. The initial velocity at the first point is 12 m/s, which is below the average velocity. Given the uniform acceleration, the velocity at the second point must exceed 30 m/s, confirming the principles of kinematics. The relevant equation used in the discussion is V = D/S, where V is velocity, D is distance, and S is time.
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You seem to have calculated the car's average velocity. Is its velocity as it passes the first point above or below that average?asdek said:Problem Statement: A car uniformly accelerates along a straight road, goes the distance between two points 240 m apart in 8 s. Its velocity as it passes the first point is 12 m/s. Find the acceleration of the car and its velocity as it passes the second point.
Relevant Equations: V=D/S
240/8=30.
jbriggs444 said:You seem to have calculated the car's average velocity. Is its velocity as it passes the first point above or below that average?
I repeat:asdek said:I'm sorry I don't understand, its velocity as it passes the first point is 12m/s. I'm not sure how to find out its velocity as it passes the secont point.
Below?jbriggs444 said:I repeat:
You seem to have calculated the car's average velocity. Is its velocity as it passes the first point above or below that average?
Yes. 12 m/s is less than 30 m/s.asdek said:Below?