Two persons A and B start from the same location and walked around a square in opposite directions with constant speeds. The square has a side 60 m…

Question

Two persons A and B start from the same location and walked around a square in opposite directions with constant speeds. The square has a side 60 m. Speeds of A and B are 4 m s^(-1) and 2 m s^(-1) respectively. When will they meet for the first time?

MARKS Content Team · Accepted Answer

Correct answer: (B) 40 s. Given, speed of the persons A and B around the square = 4 m s^(-1) and 2 m s^(-1), respectively. Let the time t, after which they meet each other. So, total distance travelled by the person A and B in time t is equal to the circumference of the square i.e., 4t+2t=4×60 6t=240 t=40 s

Two persons $A$ and $B$ start from the same location and walked around a square in opposite directions with constant speeds. The square has a side $60 m$ . Speeds of $A$ and $B$ are $4 m s^{- 1}$ and $2 m s^{- 1}$ respectively. When will they meet for the first time?

Solution

More Motion In One Dimension Questions

Two persons A and B start from the same location and walked around a square in opposite directions with constant speeds. The square has a side 60 m. Speeds of A and B are 4 m s-1 and 2 m s-1 respectively. When will they meet for the first time?

Solution

More Motion In One Dimension Questions

Two persons $A$ and $B$ start from the same location and walked around a square in opposite directions with constant speeds. The square has a side $60 m$ . Speeds of $A$ and $B$ are $4 m s^{- 1}$ and $2 m s^{- 1}$ respectively. When will they meet for the first time?