Physics, Relative velocity Class 11.
Help, stuck with a basic concept ik, can someone help me understand it?
28 Replies
@Gyro Gearloose
Note for OP
+solved @user to close the thread when your doubt is solved. Mention the user who helped you solve the doubt. This will be added to their stats.whats the q
not a q, not able to understand it, like whats the purpose, idk how to explain my doubt ig
oh then what
what does it help in? its very confusing
What is it? What is the concept/concept application?
all these type of sums
Diagram for A
yea its kinda hard to understand what role here relative velocity plays here.
use concept of relative velocity to and then motion equations to solve... basically B needs to overtake A without crashing into C when both are equally placed from A at 1km
wht does it do? like what is its use?
VelB = 56-36, VelC = 56+36
v^2=u^2 + 2as
Oh basically rel velocity means that from ground frame we understand the velocity of a certain body. Because A is also moving and the whole system surrounds it, we need to calculate velocities of B and C wrt to A
yea yea ik that anol, but its still confusing
whenever you see everything moving, you use rel velocity. imagine yourself sitting inside car a. will car c seem like its coming towards/going away from you really fast or really slow? the same for car b
wheres this from?
Imagine urself in the car and you will figure out how to relate velocity. Dont try to mug up stuff, that uk when im moving away then i add and so on and so forth... visualize.
SL Arora
dude i cant understand its basics like what is its use there
bro thats to solve the sum, what else is the use
try a few basic problems and then jump in into these, do those easy swimmer-river ones from hcv itll boost ur confidence
ohk thanks will try
when frame of reference not at rest
suppose you are walking on a road with a velocity of 10m/s and a car is approaching you at a speed of 20m/s and at this instant the car is 300m away. Calculate the time it takes for the car to hit you.
Now in this scenario you can have 2 mdthods to calculate this:
1) In ground frame consider you meet at some x dist from your original posn and then solve weird quadratic equations to get the ans.
2) In your frame of reference. For you, your own velocity is 0 and the car doesn't approach you with a speed of 20m/s but with a speed of 30m/s and it now has to travel 300m to meet you.
The second method is better as you can solve it easily with equations of motion and the ans is (300m)/(30m/s) = 10s.
awesome explanation @@Ameya² !
okk thanks
thank you sir! 🙏🏻
+solved @@Ameya² @trin @KaiZên @glenn quagmire
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