A car speeding up from rest to constant accel. travels 130m in 6s. How far in

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SUMMARY

A car accelerates from rest and covers 100 meters in 6.04 seconds. To determine the distance traveled in the first 2.37 seconds, the user initially calculated an average velocity of 16.5 m/s but incorrectly applied it to find the distance. The correct approach involves using kinematic equations, specifically the formula for uniformly accelerated motion, rather than simply multiplying average velocity by time. Additionally, the user was advised to graph velocity versus time to better understand the relationship between these variables.

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  • Understanding of kinematic equations for uniformly accelerated motion
  • Basic knowledge of graphing techniques, specifically velocity vs. time graphs
  • Familiarity with calculating average velocity
  • Concept of acceleration and its impact on distance traveled
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  • Study the kinematic equation: d = vt + 0.5at²
  • Learn how to graph velocity vs. time for uniformly accelerated motion
  • Explore the concept of acceleration and its calculation from velocity changes
  • Practice solving problems involving distance, time, and acceleration
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Students in physics, automotive engineers, and anyone interested in understanding motion dynamics and kinematics.

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A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?

So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but this is wrong. I was also told to try drawing a graph of velocity vs time...but how can I graph numbers such as these (high and low #'s with decimals) If I can't graph it I can't find the slope or y intercept...so I'm lost
 
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Draw a graph of velocity vs. time.
Distance travelled=velocity x time
 

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