Car's maximum velocity in a straight then deceleration

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SUMMARY

The discussion focuses on calculating the maximum velocity a car can achieve while navigating a 500m straight section of a track, following a corner at a velocity of 31.9 m/s. Key parameters include a friction coefficient of k=1.6, a car mass of 750 kg, and an engine power of 350 kW. The braking equation is established as a = -gk, where g is the acceleration due to gravity (9.8 m/s²). Participants suggest using constant acceleration equations to determine the distances required for both acceleration to maximum speed and deceleration back to 31.9 m/s, ensuring the total distance does not exceed 500m.

PREREQUISITES
  • Understanding of kinematics and constant acceleration equations
  • Knowledge of friction coefficients and their impact on vehicle dynamics
  • Familiarity with basic physics concepts such as mass, force, and acceleration
  • Ability to apply equations of motion in practical scenarios
NEXT STEPS
  • Study the equations of motion for constant acceleration and deceleration
  • Research the effects of friction on vehicle performance, particularly in racing contexts
  • Learn how to calculate braking distances based on initial speed and deceleration
  • Explore vehicle dynamics simulations to visualize acceleration and braking scenarios
USEFUL FOR

Automotive engineers, race car drivers, physics students, and anyone involved in vehicle performance optimization will benefit from this discussion.

leathley
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We are doing a project to find the quickest time round a given track in a car.
friction: k=1.6
mass of car m=750kg
power of engine w=350kw
g=9.8
distance x=500m

After a corner, the car comes out at velocity=31.9m/s
There is a 500m straight next but have to finish it at 31.9m/s for next corner.
I need to know the fastest way to complete the 500m.
how can i work out the maximum velocity i can go and when to start braking?
i think we found the braking eqaution was a=-gk but we are basically really stuck!
any help would be greatly appreciated!
thanks
 
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Welcome to PF!


Hi leathley! Welcome to PF! :smile:

Call the maximum speed v …

then use the appropriate constant acceleration equation twice, once to find the distance needed to accelerate up to that speed, and once to find the distance needed to decelerate under braking back to 31.9m/s …

then choose v so that the distances add up to 500m :wink:
 
distance needed to get to vmax, s = (vmax^2 -vi^2)/2a or what? can't just do the reverse for the other distance because s1 + s2 will cancel each other out.
 

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