What is the Minimum Time for Acceleration in an Auto Engine Power Problem?

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The discussion focuses on calculating the minimum time required for a car with a mass of 1230 kg to accelerate from rest to 29.1 m/s, given a maximum engine power of 92.4 kW. The theoretical calculation yields a time of 5.64 seconds, based on the work-energy theorem and power formula. However, the actual test time recorded is 12.3 seconds, indicating a discrepancy. This difference is attributed to the unaccounted friction force affecting the acceleration. Understanding these factors is crucial for accurate performance assessments in automotive engineering.
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Homework Statement


An auto manufacturer reports that the maximum power delivered by the engine of a car of mass 1230 kg is 92.4 kW. Find the minimum time in which the car could accelerate from rest to 29.1 m/s. In a test the time to do this is found to be 12.3 s. Account for the difference in this time.

Homework Equations


P = W/t
Work-KE theorem

The Attempt at a Solution


t = \frac{W}{P}
= \frac{.5(1230)(29.1)^{2}}{92.4(10^{3})}
= 5.64s

The difference in time is due to not accounting for friction force.
 
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Well done: did you have a question about this problem?
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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