   Chapter 5, Problem 57P

Chapter
Section
Textbook Problem

A 1.50 × 103-kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that, air resistance remains constant at 4.00 × 102 N during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at t = 12.0 s, just before the car stops accelerating.

(a)

To determine
the average power developed by the engine.

Explanation

Section1:

To determine: The engine force.

Answer: The engine force is 2.65×103N .

Explanation:

Given Info:

The mass of the car is 1.50×103kg .

The acceleration of the car is 18.0m/s .

The time of acceleration of the car is 12.0s .

The air resistance is 4×102N .

The acceleration of the car is,

a=vv0t

• v0 is initial velocity
• v is the final velocity
• t is time of acceleration

Formula to calculate the engine force is,

Fengine=Fair+ma

• Fair is the air resistance
• m is the mass of the car

On re-arranging,

Fengine=Fair+m(vv0t)

Substitute 4×102N for Fair , 1.50×103kg for m, 18.0m/s for v, zero for v0 and 12.0s for t to find the engine force,

Fengine=400N+(1.50×103kg)(18

(b)

To determine
The instantaneous power output of the engine just before the car stops accelerating.

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