Acceleration of object - how much time does it take

  • Thread starter Thread starter lumtesm
  • Start date Start date
  • Tags Tags
    Acceleration Time
AI Thread Summary
To calculate the distance an object needs to reach 45 mph in 3.7 seconds, first determine the acceleration using the formula a = (v_f - v_i) / Δt. For 45 mph, which equals 66 ft/s, the acceleration is approximately 17.84 ft/s². Using the distance formula x = 0.5 * a * t², the distance required is about 122.1 feet. This calculation is relevant for racing go-karts and similar applications.
lumtesm
Messages
1
Reaction score
0
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
 
Physics news on Phys.org
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
x_f = {1 \over 2} a_x t^2


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

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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top