Acceleration of object - how much time does it take

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SUMMARY

The discussion focuses on calculating the distance required for an object to accelerate from 0 mph to 45 mph within a time frame of 3.7 seconds. The acceleration is determined using the formula a = (v_f - v_i) / Δt, resulting in an acceleration of 17.84 ft/s². Subsequently, the distance is calculated using the formula x_f = 0.5 * a * t², yielding a total distance of 122.1 feet. This calculation is particularly relevant for go-kart racing enthusiasts seeking to optimize their performance.

PREREQUISITES
  • Understanding of basic physics concepts, specifically acceleration and distance calculations.
  • Familiarity with the formula for acceleration: a = (v_f - v_i) / Δt.
  • Knowledge of unit conversions between miles per hour and feet per second.
  • Ability to apply kinematic equations for uniformly accelerated motion.
NEXT STEPS
  • Research the implications of acceleration on go-kart performance and handling.
  • Learn about kinematic equations and their applications in racing scenarios.
  • Explore unit conversion techniques between different measurement systems (imperial and metric).
  • Investigate advanced acceleration calculations for varying conditions, such as friction and incline.
USEFUL FOR

This discussion is beneficial for go-kart racers, physics students, and automotive engineers interested in performance optimization and acceleration calculations.

lumtesm
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hey

does anyone know how to calcuate the distance it would take for an object to reach 45mph starting from 0mph with a time limit of 3.7 seconds?

its not a homework question, i acctually race go karts and was curious if there was a way to calculate that

thanks for your help
Mike
 
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lumtesm said:
hey

does anyone know how to calcuate the distance it would take for an object to reach 45mph starting from 0mph with a time limit of 3.7 seconds?

its not a homework question, i acctually race go karts and was curious if there was a way to calculate that

thanks for your help
Mike

EDIT OOPS! I was sure you had asked about the *acceleration*!
Once you have the acceleration, you may calculate the distance using
[tex]x_f = {1 \over 2} a_x t^2[/tex]


Of course. The acceleration is simply [itex]{v_{x,f} - v_{x,i} \over \Delta t }[/itex].

You have to be careful with units. I am not sure what units people use quote acceleration of cars in the non-metric system, but it could be in mph/s (miles per hour per second) or miles per hour squared or miles per second squared. I have a hunch that the first would be the standard one (mph per second) in which case you would just divide 45 mph by 3.7 seconds. (In the metric system, it's of course given in m/s^2, meter per second squared).

Patrick
 
Last edited:
45 mph = 66 ft/s, use ft/s in the above formula.
Therefore, acc. = 66 ft/s / 3.7 s = 17.84 ft/s^2 = a.

So the distance, x, using the formula = 0.5 (17.84) (3.7)^2 = 122.1 feet.
 

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