Question

A 400 hp engine in a 1,600 kg car applies maximum force for 2 seconds to accelerate the car onto the

Answer 1

Answer:

I will assume that “maximum force” implies the constant application of power  P  = 400 hp (international) to accelerating the vehicle. The force will therefore vary with speed as the vehicle accelerates. I will also assume that all engine energy goes into accelerating the vehicle, rather than rotating elements like its wheels.

In this case the 400 hp (equivalent to 298,280 watts) is applied for time  t  = 2 seconds. Therefore the kinetic energy of the vehicle is increased by:

ΔKE=Pt=(298,280)(2)=596,560  joules.

The initial kinetic energy is:

KEinitial=12mv2

=(0.5)(1600)(82)=51,200  joules.

Therefore final kinetic energy is:

KEfinal=KEinitial+ΔKE

=51,200+596,560

=647,760  joules

Therefore final vehicle velocity can be found:

KEfinal=12mv2

v=2KEfinalm−−−−−−−−√

=(2)(647,760)1600−−−−−−−−−−−√

= 28.455 m/s

Explanation: