a drives his car at 360 m/s. moving ahead for some hours his breaks down. so he takes 20 seconds to fix it. mean while he notices that another car which was 400 m back is now 200m ahead of his car. what is the speed of the car?
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The velocity of car 2 relative to car 1 at time t=t is
v=20-at=20–2t…………….(1)
In order to overtake car 2 , car 1 has to cover the relative distance of 10m with velocity v given by equation (1).
The relative distance is 5+5=10m.
Therefore ,10 =integral front=0 to to of (20–2t) dt= 20t-t^2
OR
t^o2 -20to+10= 0…………….(2)
Then, to=[20 +_sqrt(400–40)]/2 or
to=10 -9.49=0.51s or19.49s
During time 0.51s ,car 2 travels the road distance ,
d=20(0.51)=10.2m
Now, look at time t=19.49s
At this time car 1 moves with velocity v1=38.98 m/s . The distance it covers is d1=(1/2)(2)(19.49)^2=379.9m.
In this time car 2 moves through distance, d2=(19.49)(20)=389.8m.
So ,at this time car 2 again over takes the car 1.
v=20-at=20–2t…………….(1)
In order to overtake car 2 , car 1 has to cover the relative distance of 10m with velocity v given by equation (1).
The relative distance is 5+5=10m.
Therefore ,10 =integral front=0 to to of (20–2t) dt= 20t-t^2
OR
t^o2 -20to+10= 0…………….(2)
Then, to=[20 +_sqrt(400–40)]/2 or
to=10 -9.49=0.51s or19.49s
During time 0.51s ,car 2 travels the road distance ,
d=20(0.51)=10.2m
Now, look at time t=19.49s
At this time car 1 moves with velocity v1=38.98 m/s . The distance it covers is d1=(1/2)(2)(19.49)^2=379.9m.
In this time car 2 moves through distance, d2=(19.49)(20)=389.8m.
So ,at this time car 2 again over takes the car 1.
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