Mechanics | Q1 THE COORDINATES OF A MOVING PARTICLE AT A TIME T,-askIITians

tapish khari Grade: 12

         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

Answers : (4)

AskiitianExpert Shine

10 Points

										Hi
Ans 1 :Find the velocity in each direction seperately: 
for X direction V_x= ðx/ðt = 5x10cos10T
for Y direction V_y= ðy/ðt = 5 x 10 ( - sin10T)
For calculating speed , we need to calculate ( V_x²+ V_y²)^1/2.
Hence answer is 50.

8 years ago

AskiitianExpert Shine

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)

one year ago

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