Question

a particle moves according to the position function x(t)=ct^2+bt, with c = 2m/s and b= -5m/s. Find the acceleration and velocity when the particle is at the origin.

Answer 1

Answer on Question #46835, Physics, Mechanics | Kinematics | Dynamics

A particle moves according to the position function $x(t) = ct^2 + bt$, with $c = 2\, \text{m/s}$ and $b = -5\, \text{m/s}$. Find the acceleration and velocity when the particle is at the origin.

Solution:

The kinematics equation is:

$$x \ t = x_0 + v_0 t + \frac{a t^2}{2}$$

where $x_0$ is the initial position, $v_0$ is the initial velocity and $a$ is the acceleration.

From given we have

$$x \ t = c t^2 + b t = 2 t^2 - 5 t$$

Thus, from comparing two equations we have

$$x_0 = 0$$ $$v_0 = -5\, \text{m/s}$$ $$a = 4\, \text{m/s}^2$$

Answer: $a = 4\, \text{m/s}^2$, $v_0 = -5\, \text{m/s}$.

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