Question

Find the instantaneous velocity when x = 9 + 5t^2 after t = 3s

Answer 1

Explanation:

ans is 30

$\frac{dx}{ dt} = 10t$

t = 3 that applied v = 30

Answer 2

Answer:

$\large \implies30 \ m/sec$

Explanation:

Given :

Displacement ( x )  = 9 + $5t^2$

We have to find  instantaneous velocity at t = 3 sec

We know that

$\large v=\dfrac{dx}{dt}$

Differentiate x with respect to t

$\large v=\dfrac{d(9+5t^2)}{dt}\\\\\\\large \text{Applying power rule i.e. x^n=n.x^{n-1} }\\\\\\\large \text{ and we know differentiation of constant is 0}\\\\\\\large \implies\dfrac{d}{dt}\left(5t^2+9\right) \\\\\\\large \implies2\times5t^{2-1}+0\\\\\\\large \implies10t\\\\\\\large \impliesNow \ put \ t=3\\\\\\\large \implies10\times3 \ m/sec\\\\\\\large \implies30 \ m/sec$

Thus we get instantaneous velocity = 30 m / sec.