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.
$(i)$ $v=7 m s ^{-1}$
$(ii)$ $\quad a=\frac{v-u}{t}=\frac{14-2}{6}=2 m s ^{-2}$
$(iii)$ $S =$ Area under the graph
$=$ Area of the trapezium $ABCD$
$=\frac{1}{2}[ AD + BC] \times CD$
$=\frac{1}{2}[6+14] \times 4=40 m$
Derive following equations for a uniformly accelerated motion
$(i)$ $v=u+a t$
$(ii)$ $S=u t+1 / 2 a t^{2}$
$(iii)$ $v^{2}-u^{2}=2 a S,$ where symbols have their usual meaning.
A body is thrown vertically upward with velocity $u$, the greatest height $h$ to which it will rise is,
Can a body have constant speed and still be accelerating ?
What can you conclude about the motion of a body depicted by the velocity$-$time graphs $(i),(i i)$ and $(i i i)$ given below
There is an argument about uniform acceleration between $Mr$ $X$ and $Mr$ $Y.$ $Mr$ $X$ says "acceleration means that farther you go faster you go". $Mr$ $Y$ says "acceleration means that longer you go the faster you go". Whose statement is correct ?