A car of mass 942.4 kg accelerates from rest with a constant power output of 140.5 hp. Neglecting air resistance, what is the speed of the car after 4.55 s?

Short Answer

Expert verified
Answer: The final speed of the car after 4.55 seconds is approximately 30.49 m/s.

Step by step solution

01

Convert the power to Watts

Firstly, we need to convert the power from horsepower (hp) to Watts (W), since SI-units are preferred. We know that 1 hp = 746 Watts. So, the power output in Watts: Power (in Watts) = 140.5 hp * 746 W/hp = 104863.7 W
02

Find the work done

To find the work done, we use the equation: Work = Power * Time. In this case, the power is 104863.7 W, and the time is 4.55 seconds. Work Done = 104863.7 W * 4.55 s = 477125.835 J (Joules)
03

Use the work-energy theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. Since the car starts from rest, its initial kinetic energy is 0. Therefore, the final kinetic energy is equal to the work done. Final Kinetic Energy = 477125.835 J
04

Find the final velocity

To find the final velocity, we use the kinetic energy formula, which is: KE = (1/2) * m * v^2, where m is the mass, v is the velocity, and KE is the kinetic energy. We have the mass (942.4 kg), and the final kinetic energy (477125.835 J). Now, we will solve for v: (1/2) * m * v^2 = Final Kinetic Energy v^2 = (2 * Final Kinetic Energy) / m v = sqrt((2 * 477125.835 J) / 942.4 kg) After calculating, we get: v = 30.49 m/s The speed of the car after 4.55 seconds is approximately 30.49 m/s.

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Most popular questions from this chapter

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is \(1143.5 \mathrm{~kg}\). The coefficient of static friction between the bed of the truck and the bottom sack in the stack is \(0.372,\) and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to \(56.6 \mathrm{mph}\) in \(22.9 \mathrm{~s}\). The stack of sacks is \(1 \mathrm{~m}\) from the end of the truck bed. Does the stack slide on the truck bed? The coefficient of kinetic friction between the bottom sack and the truck bed is \(0.257 .\) What is the work done on the stack by the force of friction between the stack and the bed of the truck?

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