QueenTima
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A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show all your work, including the equation used, given and unknown quantities, and any algebra required. Make sure your answer has the correct number of significant figures.

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Answer:

Explanation:

[tex]v^2-u^2=2as\\40^2-10^2=2a*125\\1600-100=250a\\a=\frac{1500}{250}=6~m/s^2\\v=u+at\\40=10+6t\\6t=30\\ t=\frac{30}{6}=5 ~s[/tex]

onyebuchinnaji

The time of motion of the car to the given distance is 5 seconds.

The given parameters;

  • initial velocity of the car, u = 10 m/s
  • final velocity of the car, v = 40 ms/
  • distance traveled by the car, s = 125 m

The time of motion of the car to the given distance is determined by applying average velocity formula as shown below;

[tex]s = (\frac{u+ v}{2} )t\\\\125 = (\frac{10 + 40}{2} )t\\\\125 = 25t\\\\t = \frac{125}{25} \\\\t = 5 \ s[/tex]

Thus, the time of motion of the car to the given distance is 5 seconds.

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