Question

s = 10t + 4t^2 find acceleration ​

Answer 1

Solution :-

Given the expression of s

s = 10t + 4t²

Now as we know

$a = \dfrac{dv}{dt}$

$v = \dfrac{ds}{dt}$

So first of all we should know about Differentiation.

▪️differentiation of a constant = 0

$\bullet \dfrac{d(t^{n})}{dt} = (n)t^{n-1}$

$\bullet \dfrac{d(kt^n)}{dt} = k \times \dfrac{d(t^n)}{dt}$

$\bullet \dfrac{d(t^{a} + t^{b})}{dt} = \dfrac{d(t^{a})}{dt} + \dfrac{d(t^b)}{dt}$

Now coming back to our questions :-

s = 10t + 4t²

So

$v = \dfrac{ds}{dt}$

$v = \dfrac{d(10t + 4t^2)}{dt}$

$v = 10 \dfrac{dt}{dt} + 4 \dfrac{dt^2}{dt}$

$v = 10 \times t^0 + 4 \times 2 \times t^1$

$v = 10 + 8t$

$v = 8t + 10$

Now Acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d(8t + 10)}{dt}$

$a = 8\dfrac{dt}{dt} + \dfrac{d(10)}{dt}$

$a = 8 + 0$

So Acceleration

$\Huge{\boxed{\sf{a = 8 \: m/s^2 }}}$