Question

- Velocity of an object is given by V= (4x+5)^2, find
its acceleration at x=1.​

Answer 1

Given,

v = (4x+5)²

To find,

acceleration at x = 1

we know, that differentiation of velocity gives acceleration.

so,

Here,

v = (4x+5)²

=>$\frac{d(v)}{dx}$

=>$\frac{d( {4x + 5)}^{2} }{dx}$

=>2(4x+5)²-¹

=>2(4x+5)

=>8x+10

putting, the value of x = 1

8x+10

=>8×1+10

=>8+10

=>18

So, acceleration at x= 1 is 18m/s²

Cheers!