The speed of the particle is given by?

Question

The co-ordinates of a moving particle at any time t are given by x = ct^2 and y = bt^2. The speed of the particle is given by:

A)2t(c+b)

B)2t√(c^2 - b^2)

C)t√(c^2 + b^2)

D) 2t√(c^2 + b^2)

Please explain the method?

Answer 1

Answer C)

Use pythagoras to find the location of the particle.

ie...  location = (x^2 + y^2)^1/2

For speed, divide location by time..

speed = ((x^2 + y^2)^1/2) /t

then simplify using algebra

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Thanks Somu... you are right... I am wrong.. Sorry F.A.M.

Answer 2

X-component of velocity Vx = dx/dt = 2ct

Y-component of velocity Vy = dy/dt = 2bt

Speed of the particle = √(Vx^2 + Vy^2) = √{(2ct)^2 + (2bt)^2}

= 2t√(c^2 + b^2)

Ans: D

UV,

What you have done is to find average speed between time 0 and t. But what is required is to find speed at a given time t.