A point moves rectilinearly in one direction. Fig. 1.1 shows the distance s traversed by the point as a function of the time Using the plot find:
(a) the average velocity of the point during the time of motion;
(b) the maximum velocity;
(c) the time moment t0 at which the instantaneous velocity is equal to the mean velocity averaged over the first t0 seconds.
Solution:
(a) Sought average velocity
\( <v>=\frac{s}{t}=\frac{200\,cm}{20\,s}=10\,cm/s \)
(b) For the maximum velocity, \( \frac{ds}{dt} \) should be maximum. From the figure \( \frac{ds}{dt} \) is maximum for all points on the line ac, thus the sought maximum velocity becomes average velocity for the line ac and is equal to:
\( \frac{bc}{ab}=\frac{100\,cm}{4\,s}=25\,cm/s \)
(c) Time t0 should be such that corresponding to it the slope \( \frac{ds}{dt} \) should pass through the point O (origin), to satisfy the ralationship \( \frac{ds}{dt}=\frac{s}{{{t}_{0}}} \). From figure the tangent at point d passes through the origin and thus corresponding time \( t={{t}_{0}}=16\,s \).
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