Solved Problem on Kinematics
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The table below describes the velocities of a particle moving in a given reference frame.

 t (2) 0 1 2 3 4 5 6 v (m/s) 9 6 3 0 -3 -6 -9

from the table find:
a) The initial speed of the particle;
b) The instant when the particle changes its direction;
c) The average acceleration of the particle between the instants 1 s and 2 s;
d) The average acceleration of the particle between the instants 5 s and 6 s;

Solution

a) From the table, we have that for t=0 the initial velocity is
$\bbox[#FFCCCC,10px] {v_{0}=9\;\text{m/s}}$

b) At time t=2 s the velocity is positive (v>0), at time t=4 s velocity is negative (v<0), the particle changes its direction when velocity is zero at t=3 s.

c) Using the expression for the average acceleration
$\bbox[#99CCFF,10px] {a_{\text{m}}=\frac{\Delta v}{\Delta t}=\frac{v_{\text{f}}-v_{\text{i}}}{t_{\;\text{f}}-t_{\text{i}}}}$
substituting the values in the table, we have
$a_{\text{m}}=\frac{3-6}{2-1}\\ a_{\text{m}}=-\frac{3}{1}$
$\bbox[#FFCCCC,10px] {a_{\text{m}}=-3\;\text{m/s}^{2}}$

d) Using the expression for the average acceleration again, we have
$a_{\text{m}}=\frac{-9-(-6)}{6-5}\\ a_{\text{m}}=\frac{-9+6}{1}\\ a_{\text{m}}=-\frac{3}{1}$
$\bbox[#FFCCCC,10px] {a_{\text{m}}=-3\;\text{m/s}^{2}}$