Question

A mining cart is pulled up a hill at 24 km/h and then pulled back down the hill at 35 km/h through its original level. (The time required for the cart's 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?

Answer 1

Average speed of the cart will be 28.4745 km/hr

Step-by-step explanation:

Let the distance of the hill is d

Velocity of the cart when it is is pulled up to the hill is 24 km/hr

So time
`t=(d)/(24)hr`

Velocity of the cart when it is pulled down = 35 km/hr

So time taken
`t=(d)/(35)hr`

We know that average velocity is given by

`average\ velocity=(total\ distance)/(total\ time)=(2d)/((d)/(24)+(d)/(35))=28.4745kmn/hr`