Finding Work and Power in a Traffic Junction

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SUMMARY

The discussion centers on a physics problem involving a 2-ton car traveling at 45 km/h approaching a traffic junction. The driver accelerates to cross the junction before the light turns red, requiring calculations of acceleration and power. The correct power output of the engine at the moment of crossing is determined to be 82.9 kW, using the equations P = W/t and W = Fd. The initial velocity is converted to 12.5 m/s, and the frictional resistance is calculated as 1.2 kN.

PREREQUISITES
  • Understanding of kinematic equations, particularly those relating distance, initial velocity, acceleration, and time.
  • Familiarity with the concepts of force, work, and power in physics.
  • Basic knowledge of unit conversions, especially between kilometers per hour and meters per second.
  • Ability to calculate frictional forces based on mass and resistance coefficients.
NEXT STEPS
  • Study kinematic equations, specifically the equation that relates distance, initial velocity, acceleration, and time.
  • Learn about the principles of dynamics, focusing on Newton's second law of motion.
  • Explore the concept of power in physics, including its calculation in various contexts.
  • Investigate real-world applications of physics in automotive engineering, particularly in vehicle acceleration and braking systems.
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Students studying physics, particularly those focusing on mechanics, as well as automotive engineers and anyone interested in the dynamics of vehicle motion at traffic junctions.

herbgriffin
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Homework Statement


A car of 2 ton mass traveling at 45 kmph approaches a traffic junction. When the car is 100m before the signal, the driver realizes that the green light is about to turn into red in 6 seconds and hence, he accelerates the car uniformly and crosses just before the light turns red. If the frictional resistance is 0.6 kN/ton, determine the power imparted by the engine at that instant.
Answer: 82.9 kW

Homework Equations


P= W/t ; W=Fd


The Attempt at a Solution


Vi = 45 kmph = 12.5 m/s
V= 100m/6sec =16.67 m/s
and i find the acceleration by the formula of V^2= Vi^2+2ax
and i use the formula of summation of forces horizontal : -Ff + F = ma
Ff= (0.6kN/ton)(2ton) = 1.2kN
and when i find F i substitute it to the formula of power F*100m , but my answer is wrong...please help...
 
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Your calculation for the velocity would be fine if the car was moving at a constant speed. Then, covering 100m in 6 seconds would yield an accurate value for the velocity. In this case, the car has an initial velocity and is accelerating. Do you have another kinematic equation in your repertoire that relates distance covered with initial velocity, acceleration, and time?
 

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