Question

Displacement of a body is given by 3s=a+4bt*3 where a and b are constants. what is the velocity v at the end of 1 second.and here * means raised to the power

Answer 1

HEY BUDDY..!!!

HERE'S THE ANSWER..

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♠️ This kind of questions are really easy , but you have to understand this concept. So here we go!!!


♠️ As we know differentiation of displacement with respect to time gives us Velocity , Here displacement is given by [ 3 s = a + 4 . b . t^3 ] , i.e d ( s ) d t = velocity

▶️ Now differentiating it with respect to time

✔️ 3 d ( s ) / d t = d ( a + 4 . b . t^3 ) d t

=> 3 . velocity = 0 + 3 × 4 . b . t^2

⏺️ As derivative of constant is zero , and rule of derivative is [ d (x^n ) d x = n . x^( n - 1 ) ]

=> velocity = 3 × 4 . b . t^2 / 3

=> velocity = 4 . b . t^2

⏺️ Now velocity for 1 sec , we'll put 1 at place of t

=> velocity = 4 . b . 1^2

=> [ velocity = 4 b ]✔️✔️


HOPE HELPED..

JAI HIND..

:-)

Answer 2

If we differentiate displacement w.r.t. time , we get velocity.


v = 4b t^2

t = 1

=> v = 4 b