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Draw a velocity versus time graph for a body which starts to move with velocity $'u^{\prime}$ under a constant acceleration $'a'$ for time $t$. Using this graph derive an expression for distance covered $'S'$ in time $'t^{\prime}$
Solution
The graph is as shown
The area under the graph is the area of the rectangle $OACD$ plus the area of the triangle ABC on top of it as shown in figure. The rectangle has a height $u$ and a length $t$. This area is the distance travelled by the object.
Hence, $S=u t+\frac{1}{2} \times t \times(v-u)$ $….(1)$
But from the expression $v=u+a t,$ we have
at $=v-u$. Substituting in equation $(1),$ we have
$S=u t+\frac{1}{2}(a t) t=u t+\frac{1}{2} a t^{2}$
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.
