A point moves rectilinearly in on direction. Figure shows

the distance s traversed by the point as a function of the time t. Using the plot find:
(a) the average velocity of the point during the time of motion,
(b) the maximum velocity,
(c) the time moment `t_0` at which the instantaneous velocity is equal to the mean velocity averaged over the first `t_0` seconds.

Answer

(a) `10cm//s`; (b) `25cm//s`; (c) `t_0=16s`.

Explanation

(a) Sought average velocity
`lt v gt =s/t=(200cm)/(20s)=10cm//s`
(b) For the maximum velocity, `(ds)/(dt)` should be maximum. From the figure `(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:
`(bc)/(ab)=(100cm)/(4s)=25cm//s`
(c) Time `t_0` should be such that corresponding to it the slope `(ds)/(dt)` should pass through the point O (origin), to satisfy the relationship `(ds)/(dt)=s/t_0`. From figure the tangent at poind d passes through the origin and thus corresponding time `t=t_0=16s`.