An object starts a linear motion with velocity $'u^{\prime}$ and with uniform acceleration ' $a^{\prime}$, it acquires a velocity $'v^{\prime}$ in timet
$(a)$ Draw its velocity$-$time graph.
$(b)$ Obtain Ist equation of motion, $v=u+a t,$ for velocity $-$ time relation by using velocity$-$time graph.
$(c)$ A body moving with a velocity of $2\, m s ^{-1}$ acquires a velocity of $10 \,m s ^{-1}$ in $5\, s$. Find its acceleration.
An object starts a linear motion with velocity $'u^{\prime}$ and with uniform acceleration ' $a^{\prime}$, it acquires a velocity $'v^{\prime}$ in timet
$(a)$ Draw its velocity$-$time graph.
$(b)$ Obtain Ist equation of motion, $v=u+a t,$ for velocity $-$ time relation by using velocity$-$time graph.
$(c)$ A body moving with a velocity of $2\, m s ^{-1}$ acquires a velocity of $10 \,m s ^{-1}$ in $5\, s$. Find its acceleration.
$(a)$ The $v-t$ graph is as shown
$(b)$ Now, slope of the $v-t$ graph gives the acceleration of the motion.
$a=\frac{v-u}{t-0}=\frac{v-u}{t}$
Hence, $v=u+a t$
$(c)$ By using velocity - time graph
$u=2 m s ^{-1}, v=10 m s ^{-1}, t=5 s$
Using the expression $v=u+a t,$ we have
$a=\frac{10-2}{5-0}=\frac{8}{5}=1.6 m s ^{-2}$
Similar Questions
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$(ii)$ Calculate the acceleration of the lift
$(a)$ During the first two seconds.
$(b)$ Between the $3^ {r d}$ and $10^ {t h}$ second.
$(c)$ During the last two seconds.
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$(c)$ During the last two seconds.
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A body is thrown vertically upward with velocity $u$, the greatest height $h$ to which it will rise is,
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