#### Question

The position-time (*x-t*) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Figure Choose the correct entries in the brackets below;

(a) (A/B) lives closer to the school than (B/A)

(b) (A/B) starts from the school earlier than (B/A)

(c) (A/B) walks faster than (B/A)

(d) A and B reach home at the (same/different) time

(e) (A/B) overtakes (B/A) on the road (once/twice).

#### Solution 1

**(a) A** lives closer to school than **B.**

**(b) A** starts from school earlier than **B.**

**(c) B** walks faster than **A.**

**(d) A** and **B** reach home at the same time.

**(e) B** overtakes **A** once on the road.

**Explanation:**

**(a) **In the given *x*–*t* graph, it can be observed that distance OP < OQ. Hence, the distance of school from the **A’s **home is less than that from **B’s **home.

**(b) **In the given graph, it can be observed that for *x* = 0, *t* = 0 for **A**, whereas for *x* = 0, *t* has some finite value for **B**. Thus, **A** starts his journey from school earlier than **B**.

**(c) **In the given *x*–*t* graph, it can be observed that the slope of **B** is greater than that of **A**. Since the slope of the *x*–*t* graph gives the speed, a greater slope means that the speed of **B** is greater than the speed **A**.

**(d) **It is clear from the given graph that both **A** and **B** reach their respective homes at the same time.

**(e)** **B** moves later than **A** and his/her speed is greater than that of **A.** From the graph, it is clear that **B**overtakes **A** only once on the road.

#### Solution 2

(a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],

(b) A starts earlier for school than B, because t = 0 for A but for B, t has some finite time.

(c) As slope of B is greater than that of A, thus B walks faster than A.

(d) A and B reach home at the same time.

(e) At the point of intersection (i.e., X), B overtakes A on the roads once