Question

A car covers the first half of the distance between two places at the speed of 40 km per hour and other half 60 km per hour the average speed of the car in this trip is

Answer 1

Speed for first half distance(v1)= 40 km/h

Speed for second half distance(v2)= 60 km/h

Average speed for trip = 2×v1×v2/(v1+v2) = 2×40×60/(40+60) = 4800/100 = 48 km/h

Average speed of trip is 48 km/h.

Answer 2

$\underline{\large\bf{\mathfrak{Bonjour!}}}$

Let total distance travelled be "d".

Speed of the car in first half of the distance (v1) = 40 km/hr

Speed of the car in second half of the distance (v2) = 60 km/hr

Time taken to cover first half of the distance (t1) = $\frac{d}{2v1} = \frac{d}{80} hr$

Time taken to cover second half of the distance (t2) = $\frac{d}{2v2} = \frac{d}{120} hr$


$Average speed = \frac{total \: distance \: travelled}{total \: time}\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{d}{t1 + t2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{d}{ \frac{d}{80} + \frac{d}{120} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{d}{ \frac{3d + 2d}{240} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240d}{5d} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{240}{5} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =48 \: km/hr$


Therefore the average speed of the car in this trip = 48 km/hr

$\bf{\mathfrak{Hope \: this \: helps...:)}}$