(a) -1.208 m
The position of the particle at time t is given by
[tex]x(t) = x_0 + v_0 t + \frac{1}{2}at^2[/tex]
where:
[tex]x_0 = 0.250 m[/tex] is the initial position
[tex]v_0 = 0.050 m/s[/tex] is the initial velocity
[tex]a=-0.240 m/s^2[/tex] is the acceleration
Substituting into the equation t=3.70 s, we find the position after 3.70 seconds:
[tex]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[/tex]
(b) -0.838 m/s
The velocity of the particle at time t is given by:
[tex]v(t) = v_0 + at[/tex]
where
[tex]v_0 = 0.050 m/s[/tex] is the initial velocity
[tex]a=-0.240 m/s^2[/tex] is the acceleration
Substituting t = 3.70 s, we find the velocity after 3.70 seconds:
[tex]v(3.70 s) = 0.050 m/s + (-0.240 m/s^2)(3.70 s)=-0.838 m/s[/tex]
