Question

The position of a particle moving on a straight line is given by s(t)=tan(4t). What is the acceleration of the particle at time t?

Answer 1

Answer:

Acceleration of the particle is $32tan(4t)sec^{2}(4t)$

Explanation:

Given that the position of the particle which is moving on the straight line is,

$s(t)=tan4(t)$

Differentiate it with respect to t will give the velocity,

$v=\frac{ds}{dt}\\ v=\frac{d(tan4t)}{dt}\\ v=4sec^{2}(4t)$

Now again differentiate it with respect to t to get acceleration.

$a=\frac{dv}{dt}\\ a=\frac{d(4sec^{2}(4t) }{dt} \\a=8sec(4t)(4)sec(4t)tan(4t)\\a=32tan(4t)sec^{2}(4t)$

Therefore vthe acceleration of the particle is, $32tan(4t)sec^{2}(4t).$