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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