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The graph given below is the distance$-$time graph of an object.
$(i)$ Find the speed of the object during first four seconds of its journey.
$(ii)$ How long was it stationary ?
$(iii)$ Does it represent a real situation ? Justify your answer.
Solution
$(i)$ $OP$ indicates uniform speed given
$v=\frac{75-0}{4-0}=\frac{75}{4}=18.75 m s ^{-1}$
$(ii)$ $PQ$ is parallel to the time axis. It indicates that the object is stationary. It is a real situation.
$(iii)$ $QR$ shows that distance and time decreases, which is not possible. Thus, this is not a real situation.
Similar Questions
An object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time
| Time $(s)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Velocity $\left( m s ^{-1}\right)$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ |
Plot the graph.
From the graph.
$(i)$ Find the velocity of the object at the end of $2.5 sec$
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate' the distance covered in the last $4$ sec.
