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Chapter 2: Q85P (page 38)

A mining cart is pulled up a hill at 20 km/hand then pulled back down the hill at 35 km/hthrough its original level. (The time required for the cart鈥檚 reversal at the top of its climb is negligible).What is the average speed of the cart for its round trip, from its original level back to its original level?

Short Answer

Expert verified

The average speed of the cart for its round trip is 25 km/h

Step by step solution

01

Given information

v1=20km/hv2=35km/h

02

To understand the concept

This problem involves simple physical quantities speed, distance, and acceleration. When acceleration is constant one can say that speed is the ratio of distance and time.

03

Calculations for the average speed of the cart

So, the total distance the cart is being pulled will be2ykm

Time for pulling the cart up ist1=y20hrs

Time for bringing the cart down ist2=y35hrs

So, the total time is

t=t1+t2=y20+y35

The average velocity is defined as

localid="1660890442400" Averagespeed=TotaldistanceTotaltimeAveragespeed=2yy20+y35Averagespeed=2y55y700=25.45~25km/h

So, the average speed of the cart will be 25 km/h

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