Question

A particle moves along the x axis. It is initially at the position 0.250 m, moving with velocity 0.050 m/s and acceleration -0.240 m/s2. Suppose it moves with constant acceleration for 3.70 s. (a) Find the position of the particle after this time. m (b) Find its velocity at the end of this time interval. m/s

Answer 1

(a) -1.208 m

The position of the particle at time t is given by

$x(t) = x_0 + v_0 t + \frac{1}{2}at^2$

where:

$x_0 = 0.250 m$ is the initial position

$v_0 = 0.050 m/s$ is the initial velocity

$a=-0.240 m/s^2$ is the acceleration

Substituting into the equation t=3.70 s, we find the position after 3.70 seconds:

$x(3.70 s) = 0.250 m + (0.050 m/s)(3.70 s) + \frac{1}{2}(-0.240 m/s^2)(3.70 s)^2=-1.208 m$

(b) -0.838 m/s

The velocity of the particle at time t is given by:

$v(t) = v_0 + at$

where

$v_0 = 0.050 m/s$ is the initial velocity

$a=-0.240 m/s^2$ is the acceleration

Substituting t = 3.70 s, we find the velocity after 3.70 seconds:

$v(3.70 s) = 0.050 m/s + (-0.240 m/s^2)(3.70 s)=-0.838 m/s$