# Answer to Question #118229 in Mechanics | Relativity for Rashika

Question #118229
A bus starts from rest with a constant acceleration of 5 m/sec Square. at the same time a car travelling at a constant velocity of 50m/sec overtakes the Bus and passes it. find at what distance will the bus overtake the car and how fast the bus would be travelling then
1
2020-05-26T12:49:57-0400

The law of the motion with the constant acceletation (from the origin of the coordinates and from rest):

"x_{bus}(t) = \\dfrac{at^2}{2}"

where "x(t)" is the bus coordinate at time "t".

The law of the car's motion (constant speed, starts from the origin):

"x_{car}(t) = vt"

The time at which the bus will overtake the car satisfies the following consition:

"x_{bus}(t_{overtake}) = x_{car}(t_{overtake})\\\\\n\\dfrac{at_{overtake}^2}{2} = vt_{overtake} \\Rightarrow t_{overtake} = \\dfrac{2v}{a} = \\dfrac{2\\cdot 50}{5} = 20 s"

The distans of the meeting then:

"d = x_{car}(t_{overtake}) = vt_{overtake} = 50\\cdot 20 = 1000 m"

The speed of the bus at that time will be:

"v_{bus} = at_{overtake} = 5\\cdot 20 = 100 m\/s"

Answer. d = 1000 m, v = 100 m/s.

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