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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=T 2 -T WHERE TIS IN SECONDS &IS IN M\SEC. THE TIME INTERVAL FOR WHICH PARTICLE RETARDS IS T T.>1 1/2 T=1

 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



  1. T<1/2

  2. T.>1

  3. 1/2  < T<1

  4. T=1

Grade:12

4 Answers

AskiitianExpert Shine
10 Points
13 years ago

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 .

Hence answer is 50.

AskiitianExpert Shine
10 Points
13 years ago

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.

dhulipalla aravind aravind
18 Points
13 years ago

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

Swetank Gupta
35 Points
5 years ago
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) 

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