Question

Two buses A and B are at positions 50m and 100m from the origin at t=0. They start moving in the same direction simultaneously with uniform velocity of 10m/s and 5m/s. Determine the time and position at which A overtakes B.

Answer 1

Hello Dear.

Here is the 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 T
Total Distance of Bus A = 50 + 10T

Now,
Distance travelled by the Bus B in time T seconds = 5 × T
= 5T
Total 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.


Hope it helps.

Answer 2

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