Acceleration of object - how much time does it take

  • Thread starter lumtesm
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  • #1
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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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  • #2
nrqed
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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:
  • #3
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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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