Question

The position of a particle moving in a straight line at any time t is x(t) = 2t^2 + 6t + 5. What is the acceleration of the particle at t=3

Answer 1

Answer:

The acceleration of the particle at any time t

$a = 4\ m/s^{2}$

Step-by-step explanation:

Given:

The position of a particle moving in a straight line at any time t is.

$x(t) = 2t^{2} + 6t + 5$

The velocity of the particle $v = \frac{d(x)}{d(t)}$

$v = \frac{d}{d(t)}(2t^{2} + 6t + 5)$

$v= 2\times 2t + 6$

$v= 4t + 6$

So the velocity of the particle $v= (4t + 6)\ m/s$

The acceleration of the particle $a = \frac{d(v)}{d(t)}$

$a = \frac{d}{d(t)}(4t + 6)$

$a = 4\ m/s^{2}$

In this condition the acceleration does not depending upon the time, so the acceleration is constant

Therefore the acceleration of the particle at any time t $a = 4\ m/s^{2}$