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
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Expert's answer
2020-05-26T12:49:57-0400

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

xbus(t)=at22x_{bus}(t) = \dfrac{at^2}{2}

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

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


xcar(t)=vtx_{car}(t) = vt

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

xbus(tovertake)=xcar(tovertake)atovertake22=vtovertaketovertake=2va=2505=20sx_{bus}(t_{overtake}) = x_{car}(t_{overtake})\\ \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=xcar(tovertake)=vtovertake=5020=1000md = x_{car}(t_{overtake}) = vt_{overtake} = 50\cdot 20 = 1000 m

The speed of the bus at that time will be:


vbus=atovertake=520=100m/sv_{bus} = at_{overtake} = 5\cdot 20 = 100 m/s

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


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