Co-ordinates of a particle moving in a plane at any time t are given by equations x=at2 and y=bt2. What is magnitude
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distance through x-axis:
x=αt²
veocity=dx/dt = 2αt
distance through y-axis:
y=βt²
velocity, V=2βt
Make this in the vector form,as shown in the figure.
We get the resultant vector as: V= √(2αt)² + (2βt)²
=√4α²t²+4β²t²
=√4t²(α²+β²)
=2t√α²+β²
So the answer is option (b)
NOTE:
If distance (x) is given,
velocity is dx/dt
acceleration is d²x/dt²
Sorry for my bad handwrting and graph. :P
x=αt²
veocity=dx/dt = 2αt
distance through y-axis:
y=βt²
velocity, V=2βt
Make this in the vector form,as shown in the figure.
We get the resultant vector as: V= √(2αt)² + (2βt)²
=√4α²t²+4β²t²
=√4t²(α²+β²)
=2t√α²+β²
So the answer is option (b)
NOTE:
If distance (x) is given,
velocity is dx/dt
acceleration is d²x/dt²
Sorry for my bad handwrting and graph. :P
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