Question 3.2: A particle moves along the x-axis and its coordinates vary w......
A particle moves along the x-axis and its coordinates vary with time according to the relation x = t² − 2 t, where x is measured in meters and t is in seconds. The position time graph for this motion is shown in Fig. 3.6. (a) Use this graph to comment about the particle’s motion. (b) Find the displacement and the average velocity of the particle in the time intervals 0 ≤ t ≤ 1 s and 1 s ≤ t ≤ 3 s. (c) Find the velocity of the particle at t = 2 s.
(a) The particle starts from the origin of the x-axis and moves in the negative x direction for the first second. Its velocity is zero at x = −1 m when t = 1 s and then heads back in the positive x direction for t > 1 s.
(b) In the interval 0 ≤ t ≤ 1 s we have t_i = 0 and t_f = 1 s. Since x = t² − 2 t,we get x_i =t²_{i} − 2 ti = 0 and x_f = t²_{f} − 2 tf = −1 m. Thus:
The average velocity is then:
{\overline{{v}}}={\frac{\Delta x}{\Delta t}}={\frac{x_{\mathrm{f}}-x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{-1\;{\mathrm{m}}-0\;{\mathrm{m}}}{1\;{\mathrm{s}}-0\;{\mathrm{s}}}}={\frac{-1\;{\mathrm{m}}}{1\;{\mathrm{s}}}}=-1\;{\mathrm{m}}/{\mathrm{s}}According to Fig. 3.6, this value equals the slope of the straight line drawn for
this time interval.
In the interval 1 s ≤ t ≤ 3 s we have ti = 1 s and t_{f} = 3 s. Again, from x =t² − 2 t we get x_{i} = t²_{i} − 2 t_{i} = −1 m and x_{f} = t²_{f} − 2 t_{f} = 3 m. Thus:
The average velocity is then:
{\overline{{v}}}={\frac{\Delta x}{\Delta t}}={\frac{x_{\mathrm{f}}-x_{\mathrm{i}}}{t_{\mathrm{f}}-t_{\mathrm{i}}}}={\frac{3\;{\mathrm{m}}-(-1\;{\mathrm{m}})}{3\;s-1\;{\mathrm{s}}}}={\frac{4\;{\mathrm{m}}}{2 \ s}\;= 2 \ {\mathrm{m}}/{\mathrm{s}}}According to Fig. 3.6, this value equals the slope of the straight line drawn for this time interval.
(c) To find the instantaneous velocity at any time t, we use Eq. 3.5
\ v={\frac{d x}{d t}} ≡ x_{f} – x_{i} = \int\limits_{t_{i}}^{t_{f}}{v dt} ≡ Area under v-t graph (3.5)
and apply the rules of differential calculus on the coordinate x = t² − 2 t. That is:
v={\frac{d x}{d t}}={\frac{d(t^{2}-2t)}{d t}}=2\,t-2Notice that this expression gives the velocity v at any time t and indicates that v is increasing linearly with time. It tells us that v < 0 during the interval 0 ≤ t < 1 s (i.e. the particle is moving in the negative x direction), and that v = 0 at t = 1 s, and finally v > 0 for t > 1 s. When t = 2 s we use the above expression to get:
v = 2 × 2 − 2 = 2 m/s