Question

Two brothers \[X\] and \[Y\] move from home to their respective schools on the same straight path. Position-Time(x-t) graphs for two brothers are shown here. Choose the incorrect statement-
(A). School of \[X\] is closer to their home
(B). \[Y\]overtakes \[X\] on their way to school
(C). \[Y\]moves faster than\[X\]
(D). Average velocities of the two are equal between\[t=0\]and\[t={{t}_{0}}\]

Answer 1

Hint: Distance-time graphs tell us about the motion of objects travelling in a straight line. From these graphs we can calculate average speed, distance travelled and time taken. The tangent of the graph at a point gives us the speed at that particular point.

Complete answer:

The graph is plotted between distance travelled and time taken by each brother to reach their respective schools. We can observe that both brothers reached school at the same time, \[{{t}_{0}}\]. School of \[Y\] is at a greater distance than school of \[X\].
\[X\] started his journey to school at\[t=0\], but \[Y\] started at a time much later than \[X\] but still managed to reach school at the same time as \[X\].
Also the area under the curve of \[Y\] is greater than the area under the curve of \[X\]. So, the speed of \[Y\] is greater than the speed of \[X\]. So, the average velocities of both brothers will not be equal.

Since the average velocities of both bothers is unequal so the option (D) is the incorrect statement and hence the answer.

Note:
The distance time graph is a straight line which means there is a linear relationship between distance and time. Average speed= total distance travelled divided by total time taken. Area under the curve in the distance time graph gives us the average speed of objects.