Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.1)
|Veni said: (Dec 21, 2010)|
|How answer will come? Please explain.|
|Abhishek Chouksey said: (Apr 15, 2011)|
in first 10 sec it reaches only 40 m/s=1/2*10*8(uniform acc)
in next 1.25 sec at constant acc.(v=u+at) it tends to 50 m/s.
|Puru said: (Jun 3, 2011)|
|Dear, please tell me how it comes 40m/s. By which law and logic you calculate 40m/s.
|Gopala Krishna said: (Jun 30, 2011)|
|Here the car is not moving with uniform accleration upto ten seconds
area of the triangle=0.5*8*10=40m/sec(t=10)
now accleration is constant
So apply v=u+at
So total timet = 10+1.25 = 11.25
|Mani said: (Aug 25, 2011)|
|S=1/2 (8+0) 10.
Since initial velocity is zero. My guess.
|Santhosha said: (Aug 31, 2011)|
|Initially the is at rest, we consider u=40m/s, and v=50m/s then
This is the time required to travel from 40m/s velocity to 50m/s velocity. then the time required to travel the car from rest position is 10+1.25=11.25 sec.
|Ankit Bhardwaj said: (Sep 24, 2011)|
|That's very simple...
We know dat a= dv/dt
or we can say that v= integral(a dt)...............(1)
Now by graph we can see that a=(4/5)t upto the first 10 seconds
Now put v=50 in the eq.(1)
So after calculating t= 11.25 sec.
|Sekhar said: (Sep 15, 2012)|
|Actually we want calculate and differentiating function, like ankith what he was explain.|
|Wasprulez1990 said: (Oct 7, 2012)|
|The area in the graph is velocity i.e. v=at, so the total area is given i.e.final velocity 50 m/s,
So 50= area of triangle + area of rectangle.
50= (0.5 x base x height) + (width x height).
50= (0.5 x 10 x 8) + [ (t-10) x 8].
Finally t=11.25 seconds.
|Aishwarya Katale said: (Dec 2, 2013)|
|How to calculate 40? which is v?|
|Ankit Kumar said: (Oct 27, 2014)|
|Aishwarya Katale : Actually 40m/s its area under the curve. :-).|
|Manan said: (Dec 7, 2014)|
|The answer is incorrect correct answer is 6.25s.
Solution u = 0.
v = 50 m/s.
a = 8 m/s^2.
So just find t by v = u+at.
Which is 50 = 0+8t?
50 = 8t.
t = 50\8 = 6.25s.
|Hasya said: (May 17, 2016)|
|No, 11.25 is correct.
The formula v = u + at is applicable only in the case of constant acceleration.
Since up to the first 10 seconds the car is moving wth increasing acceleration area under the triangle gives final velocity at t =10 sec since initial velocity is 0.
|Kumara Swamy said: (May 31, 2016)|
|11.25 secs are the correct answer.
Equations of motion are:
v = u + at;.
s = ut + 0. 5 at^2 etc are valid when accleration is constant.
In the question, the acceleration varies with timeup to 10 sec and then it is constant so, use
v = u + at after 10 secs.
And for before 10 secs use an area under a - t graph is the velocity similarly area under v - t graph gives displacement.
You will understand this concept when you pass through basics of calculus.
|Srikanth said: (Aug 10, 2016)|
|Acceleration * time = velocity.
So,in the above graph find the area covered by line up to uniform acceleration.
i.e., Area of triangle = 1/2 * 10 * 8 = 40 m/s. And then it follows constant acceleration and the required area(rectangle) is 10. So 8 * 1.25 = 10 m/s, so a total of (40 + 10)m/s is achieved in (10 + 1.25)secs.
|Ss Ahamed Halith said: (Oct 10, 2019)|
|1. We consider the given acceleration and time (i.e a=8 ms-2 t=10s) then we find velocity for that next we pick any one choices given and add or sub for making it equal to 10 find another velocity.|
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