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```         Q1 THE COORDINATES OF A MOVING PARTICLE AT A TIME T, ARE GIVEN BY X=5SIN10T ,Y=5COS10T. THE SPEED OF THE PARTICLE IS
Q2 A PARTICLES STARTS MOVING RECTILINARLY AT TIME ,T=O SUCH THAT ITS VELOCITY CHANGES WITH TIME T ACCORDING TO THE EQUATION V=T2-T WHERE TIS IN SECONDS &IS IN M\SEC. THE TIME INTERVAL FOR WHICH PARTICLE RETARDS IS

T<1/2
T.>1
1/2  < T<1
T=1
```
8 years ago

10 Points
```										Hi
Ans 1 :Find the velocity in each direction seperately:
for X direction Vx= ðx/ðt = 5x10cos10T
for Y direction Vy = ðy/ðt = 5 x 10 ( - sin10T)
For calculating speed , we need to calculate ( Vx2 + Vy2)1/2 .
```
8 years ago
10 Points
```										Ans 2:
Find the expression for acceleration by differentiating the velocity term given, ðv/ðT= 2T -1
For particle to retard acceleration should be less than 0, therefore 2T-1< 0 , hence T<1/2 is the answer.
```
8 years ago
dhulipalla aravind aravind
18 Points
```										Q1.
ans.       vx=5sin10t
vy=scos10t
v=[vx*vx+vy*vy]1/2
=5*10[sin2+cos2]1/2
=5*10[1]1/2
=5o
```
8 years ago
Swetank Gupta
35 Points
```										differentiating above eqns to get speed dx/dt=3at^2(speed along x-axis) dy/dt=3bt^2(speed along y-axis) since speed along x and y are vector quantities and both the components are perpendicular to each other so net speed = v=√ ((speed along x)^2 + (speed along y)^2) v=√ ((3at^2)+(3bt^2)) v=3t^2√ (a^2+b^2)
```
10 months ago
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