03]Example 24. Two buses A and B are at positions 50 m and100 m from t

1.	03]Example 24. Two buses A and B are at positions 50 m and100 m from the origin at time t 0. They start moving in thesame direction simultaneously with uniform velocity of10 ms 1 and 5 ms . Determine the time and position atwhich A overtakes B.
Answer» Let the time at which the Bus A overtakes the bus B be T seconds. ∴ Distance travelled by the Bus A from the Origin in Time T = Speed× Time= 10× T= 10 TTotal Distance of Bus A = 50 + 10T Now,Distance travelled by the Bus B in time T seconds = 5× T= 5TTotal Distance of Bus B = 100 + 5T We know,Distance of Bus A after T seconds = Distance of the Bus B after T seconds. ⇒ 50 + 10T = 100 + 5T⇒ 10T - 5T = 100 - 50⇒ 5T = 50⇒ T = 50/5⇒ T = 10 seconds. Time at which the Bus A overtake the Bus B is 10 seconds. Now, Total Distance travelled by the Bus A from the origin = 50 + 10T= 50 + 10(10)= 50 + 100=150 m. ∴ Distance from the origin at which the Bus A overtake the Bus B is 150 m.

03]Example 24. Two buses A and B are at positions 50 m and100 m from the origin at time t 0. They start moving in thesame direction simultaneously with uniform velocity of10 ms 1 and 5 ms . Determine the time and position atwhich A overtakes B.

Answer»

Let the time at which the Bus A overtakes the bus B be T seconds.

∴ Distance travelled by the Bus A from the Origin in Time T = Speed× Time= 10× T= 10 TTotal Distance of Bus A = 50 + 10T

Now,Distance travelled by the Bus B in time T seconds = 5× T= 5TTotal Distance of Bus B = 100 + 5T

We know,Distance of Bus A after T seconds = Distance of the Bus B after T seconds.

⇒ 50 + 10T = 100 + 5T⇒ 10T - 5T = 100 - 50⇒ 5T = 50⇒ T = 50/5⇒ T = 10 seconds.

Time at which the Bus A overtake the Bus B is 10 seconds.

Now, Total Distance travelled by the Bus A from the origin = 50 + 10T= 50 + 10(10)= 50 + 100=150 m.

∴ Distance from the origin at which the Bus A overtake the Bus B is 150 m.

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