Question

The position of a particle as a function of time is shown in the figure. What is the particle's average velocity between time t = 0.6 s and time t = 7.2 s? What is the average speed of the particle between time t = 0.6 s and time t = 7.2 s? What is the average acceleration of the particle in the time interval between t = 0.6 s and t = 7.2 s?

Answer 1

Assume that the particle is travelling along a straight line.

Q3.

At time $t_1=0.6\,s$ position of particle is $x_1=1.2\,m$ . At time $t_2=7.2\,s$   position of particle is $x_2=2.8\,m$

Average velocity of particle between $t_1=0.6\,s$ and $t_2=7.2\,s$ is $V_{avg}=\frac{\text{displacment}}{\text{time taken}}=\frac{x_2-x_1}{t_2-t_1}=\frac{2.8-1.2}{7.2-0.6}=0.24\,m/s$

Q4.

Average speed of particle between $t_1=0.6\,s$ and $t_2=7.2\,s$ is $|\vec{V}|_{avg}=\frac{\text{distance travelled}}{\text{time taken}}=\frac{17.6}{6.6}=2.7\,m/s$

Q5.

At time $t_1=0.6\,s$ instantaneous velocity of particle is $v_1=2\,m/s$ . At time $t_2=7.2\,s$ velocity of particle is $v_2=4\,m/s$

Average acceleration of particle between $t_1=0.6\,s$ and $t_2=7.2\,s$ is $a_{avg}=\frac{v_2-v_1}{t_2-t_1}=\frac{4-2}{7.2-0.6}=0.30\,m/s^2$