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```        Q : Two particles starts moving from the same pointalong the same straight line .The first moves with constant velocity and second with const acc. a. During the time thatelapes before  the  second catches the first greatest distance b/w the particle is ?
Q: A particle is released from a tower of height 60m . The ratio of time taken to fall equal 20 m i.e t1:t2:t3 is :
Q : a lift is moving upward with an acc. 2 m/s2 .When the upward velocity of lift is 5 m/s a particle is projected vertically upward with a velocity of 3 m/s w.r.t the floor of lift by a passenger.Find the time after which it reaches the passenger again```
8 years ago

8 Points
```										A1..
The distance btwn the two particles will keep on incresing till the second particle reaches the velovity with which the 1st paricle is moving allready..
so start with the condition on the first particle to calculate the distance covered and time taken by the particle to reach the velocity "v" of the first particle... v^2 = u^2 + 2as will give u s... and v=u+at will give u the time t
with the time "t" calculated u can calculate the distance covered by the 1st one by d=v*t
now the difference of the two distances is ur answer..

```
8 years ago
anandpal shekhawat
8 Points
```										A2. Hi,
t1=(2*20/g)^1/2 =2s
t2=(t'-t1)= [(2*40/g)^1/2 - (2)] = 2[(2)^1/2 - 1]
t3=(t"-t')= [(2*60/g)^1/2 - 2(2)^1/2]= 2[(3)^1/2 - (2)^1/2]
so, t1:t2:t3 = 1:[(2)^1/2 - 1]:[(3)^1/2 - (2)^1/2]

A3. Urel. = 3m/s
Arel. = -12m/s^2
Srel. = 0
t=?
so, 0= 3*t - 1/2*12*t^2 = 3t - 6t^2
hence, t= 1/2s
```
8 years ago
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