Question

The acceleration of a particle is a constant. At t=0, the velocity of the particle is (15.8i+ 18.4j) m/s. At t= 5.1 s the velocity is 14.0j m/s.

(a) What is the particle's acceleration (in m/s^2)?
a=

(b) How do the position (in m) and velocity (in m/s) vary with time? Assume the particle is initially at the origin.
r(t)=
v(t)=

Answer 1

(a) To find the particle's acceleration, we can use the formula: a = (v2 - v1)/t where v1 is the initial velocity, v2 is the velocity at time t, and t is the time interval. Substituting the given values, we get: a = (14.0j - (15.8i + 18.4j))/5.1 s a = (-15.8i - 4.4j)/5.1 s a = (-3.1i - 0.9j) m/s^2 Therefore, the particle's acceleration is (-3.1i - 0.9j) m/s^2. (b) To find the position function, we can integrate the velocity function with respect to time: r(t) = ∫v(t) dt Integrating