Physics

Physics

Problem

An automobile of mass mm travels along a straight highway in two successive segments. • Segment I: The car ascends a section of road that is inclined at a constant angle θ\theta above the horizontal. It starts this segment with speed v0v_0, ends with the same speed v0v_0, and spends a time t1t_1 on the incline. • Segment II: Immediately after reaching level ground at the top of the incline, the car accelerates in a straight line from speed v0v_0 to speed 2v02v_0 in a time t2t_2. Assume air resistance and frictional losses are negligible and that the engine provides all necessary mechanical energy. a. Derive an expression, in symbols only, for the average mechanical power that the engine must deliver during Segment I. b. Derive an expression, in symbols only, for the average mechanical power that the engine must deliver during Segment II. c. Given that t1=t2t_1 = t_2 and that the vertical height gained in Segment I is h=v022gh = \frac{v_0^2}{2g}, determine which segment requires the greater average engine power. Provide a clear comparison supported by algebraic reasoning. d. Discuss one physical factor not considered in the analysis above that could alter the actual power required by the engine in real-world driving.
Intuitive explanation

The key insight is that energy conservation and kinematics work together here: on the incline, the car's kinetic energy stays constant while the engine fights gravity to gain potential energy; on the level road, all engine work goes into increasing kinetic energy. Use ΔPE=mgh\Delta PE = mgh for the climb and ΔKE=12m(vf2−vi2)\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) for the acceleration phase to connect the motion parameters (t1t_1, t2t_2, θ\theta) to the engine's power output and the distances traveled.

Intuitive explanation

The key insight is that energy conservation and kinematics work together here: on the incline, the car's kinetic energy stays constant while the engine fights gravity to gain potential energy; on the level road, all engine work goes into increasing kinetic energy. Use ΔPE=mgh\Delta PE = mgh for the climb and ΔKE=12m(vf2−vi2)\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) for the acceleration phase to connect the motion parameters (t1t_1, t2t_2, θ\theta) to the engine's power output and the distances traveled.

Browse the problem library