Question

A particle moves along the x axis. Its position as function of time is given by x=6.8t+8.5t^2, where t is in seconds and x is in meters. What is the acceleration as a function of time?

Answer 1

As per the question the particle moves along X-axis.

The position of the particle is a function of time

As per the question the position $x=6.8t+8.5t^2$

The velocity of the particle is given as $v=\frac{dx}{dt}$

                                                                  v=6.8+17t

The acceleration of the particle is given as

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

                                               a=17

Here the acceleration is a constant quantity which is independent of time.

From the acceleration obtained here one can infer that particle is moving with uniform acceleration .