Question

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$\sf\implies\: coordinates\: of\: moving\: particle\: at\: any \: time\:'t'\: are\: given\: by\: x\:=\:\ \alpha\:t^{3} \:and\: y\:=\beta\: t^{3}\:\\\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ the\: speed\: of\: particle\: at \: time\:$

answer :
$3t {}^{2} \sqrt{ \alpha {}^{2} + \beta {}^{2} }$

Answer 1

Answer:

Since it is a coordinate system, x = αt³ i and y = βt³ j

⇒ Velocity ( v ) = dx /dt + dy / dt

dx / dt = d ( αt³ i ) / dt = 3 αt² i

dy / dt = d ( βt³ j ) / dt = 3 βt² j

⇒ Velocity ( v ) = 3t² ( α i + β j )

Now we are required to find the speed of the particle, which is magnitude of Velocity

⇒ | v | = Speed

⇒ | v | = √ [( dx / dt )² + ( dy / dt )²]

⇒ Speed = √ [ ( 3αt² )² + ( 3βt² )² ]

⇒ Speed = √ [ 9α²t⁴ + 9β²t⁴ ]

⇒ Speed = √ 9t⁴ [ α² + β² ]

⇒ Speed = 3t² √ ( α² + β² )     [ Taking Square root for 9t⁴ ]

This is the required answer.

Answer 2

Explanation:

Answer:

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