Question

A car makes a trip due north for three-fourths of the time and due south one-fourth of the time. The average northward velocity has a magnitude of 35 m/s, and the average southward velocity has a magnitude of 41 m/s. Taking northward to be the positive direction, what is the average velocity for the trip?

Answer 1

Answer:16 m/s

Explanation:

Given

Car trip to north for three-fourth of the time

and due to south one-fourth of time

Average northwards velocity is 35 m/s

Average southwards velocity is 41 m/s

Suppose car has moved first in north direction for $\frac{3t}{4}$ time with 35 m/s

Distance traveled$=\frac{105t}{4} m$

Now car starts moving in south thus

Distance traveled $=\frac{41t}{4} m$

Therefore net displacement is $=\frac{105t}{4}-\frac{41t}{4}=\frac{64t}{4}=16t$

therefore average velocity$=\frac{Displacement}{Time}$

$V_{avg}=\frac{16t}{t}=16 m/s$